Spring 2015)Write balanced “molecular” equations (ME) and net ionic equations … ·...

1. (Spring 2015)Write balanced “molecular” equations (ME) and net ionic equations (NIE) for the following reactions, all of which involve HBr(aq) . NOTE: Points will be deducted for missing or incorrect states in your final equations, and missing or incorrect charges for ions in the NIE.

A. HBr(aq) + Sn(s)

ME: 2HBr(aq) + Sn(s) SnBr2(aq) + H2(g)

IE: 2H+ + 2Br- + Sn(s) Sn2+ + 2Br- + H2(g)

NIE: 2H+(aq) + Sn(s) Sn2+(aq) + H2(g) B. HBr(aq) + Pb(NO3)2(aq)

ME: 2HBr(aq) + Pb(NO3)2(aq) 2HNO3(aq) + PbBr2(s)

IE: 2H + 2Br- + Pb2+ + 2NO3- 2H+ + 2NO3

- + PbBr2(s)

NIE: Pb2+(aq) + 2Br-(aq) PbBr2(s) C. HBr(aq) + KC2H3O2(aq)

ME: HBr(aq) + KC2H3O2(aq) HC2H3O2(aq) + KBr(aq)

IE: H+ + Br- + K+ + C2H3O2- HC2H3O2 + K+ + Br-

NIE: H+(aq) + C2H3O2-(aq) HC2H3O2(aq)

NOTE: HC2H3O2(aq) is a weak acid, so it is only partially ionized.

2. (Fall 2012) Write balanced molecular equations (ME) and net ionic equations (NIE) for the following reactions. Be sure to include the correct states in your final equations and charges for ions in the NIE. If you have written several equations, draw a box around the one you want to be graded. Write MEs for all reactions, but if no reaction actually occurs (i.e. everything cancels in the NIE), then write “no reaction” for the NIE.

A. HCl(aq) + NaNO2(aq)

ME: HCl(aq) + NaNO2(aq) HNO2(aq) + NaCl(aq)

IE: H + Cl + Na + NO2 HNO2(aq) + Na + Cl

Note that HNO2(aq) is a weak acid so it does not dissociate.

NIE: H(aq) + NO2(aq) HNO2(aq)

B. HCl(aq) + Al(s)

ME: 6HCl(aq) + 2Al(s) 3H2(g) + 2AlCl3(aq)

IE: 6H+ + 6Cl- + 2Al(s) 3H2(g) + 2Al3 + 6Cl-

NIE: 6H+ + 2Al(s) 3H2(g) + 2Al3+ C. HCl(aq) + Pb(NO3)2(aq)

ME: 2HCl(aq) + Pb(NO3)2(aq) 2HNO3(aq) + PbCl2(s)

IE: 2H + 2Cl + Pb2 + 2NO3 2H + 2NO3

+ PbCl2(s)

NIE: Pb2(aq) + 2Cl(aq) PbCl2(s)

Gen Chem I Homework Solutions, Unit II – page 2

D. The following chemical equations are balanced, but they do not correspond to actual chemical equations. Explain why no reaction occurs if the reactants are combined.

1. HBr(aq) + KNO3(aq) HNO3(aq) + KBr(aq) Both of the reactants and both of the products are strong electrolytes, that is, they are

essentially totally ionized. Therefore there is no product (e.g. a precipitate, a weak electrolyte) that indicates a new chemical species has formed; hence, no reaction.

2. HBr(aq) + Ag(s) AgBr(s) + 1/2H2(g) Silver is below hydrogen on the activity series – it is less “active” than hydrogen – so it will not

transfer it’s electrons to H+(aq). 3. (Fall 2015) Write balanced molecular equations (ME) and net ionic equations (NIE) for the following

reactions. Be sure to include the correct states in your final equations and charges for ions in the NIE.

A. CuCl2(aq) + Pb(s) ME: CuCl2(aq) + Pb(s) --> Cu(s) + PbCl2(s) IE: Cu2+ + 2Cl- + Pb(s) --> Cu(s) + Pb2+ + 2Cl- NIE: Cu2+(aq) + 2Cl-(aq) + Pb(s) --> Cu(s) + PbCl2(s) B. K2SO4(aq) + CaCl2(aq) ME: K2SO4(aq) + CaCl2(aq) --> 2KCl(aq) + CaSO4(s) IE: 2K+ + SO4

2- + Ca2+ + 2Cl- --> 2K+ + 2Cl- CaSO4(s) NIE: Ca2+(aq) + SO4

2-(aq) --> CaSO4(s) C. KF(aq) + HI(aq) ME: KF(aq) + HI(aq) --> KI(aq) + HF(aq) IE: K+ + F- + H+ + I- --> K+ + I- + HF(aq) NIE: H+(aq) + F-(aq) --> HF(aq) 4. (Spring 2015) On the last exam, we considered potential applications of nanoparticles based on calcium

carbonate, CaCO3(s). In Chapter 4 we saw that dissolving carbon dioxide in water forms carbonic acid, H2CO3(aq), a weak acid. We demonstrated in class that a solution of sodium hydroxide reacts with the carbonic acid formed by adding “dry” ice, CO2(s), to the solution, and we followed the reaction using a “universal” indicator.

H2CO3(aq) + 2NaOH(aq) Na2CO3(aq) + 2H2O(l)

Cancer cells are more acidic than noncancerous tissue. Because carbonate is the anion of a weak acid, carbonate compounds are basic, and because nanoparticles tend to accumulate in cancer cells, it is hoped that nanoparticles based on calcium carbonate might be used to treat cancer.

One way to determine the amount of calcium carbonate in a sample is to react it with a strong acid, such as... well... HBr(aq).

2HBr(aq) + CaCO3(s) H2CO3(aq) + CaBr2(aq)


2

hydroselenic

4

1

1

2

2

4

6

selenic

selenous

A sample of CaCO3 nanoparticles was reacted with HBr(aq) according to the chemical equation shown above. It took 267 mL of 0.114 M HBr(aq) to react with all of the calcium carbonate. How many grams of CaCO3 were contained in the sample? (The molar mass of CaCO3 is 100.09 g/mol.)

We start with the “known quantity”: 267 mL of HBr(aq).

×

1 mol CaCO 100.09 g CaCO1 L 0.114 mol HBr267 mL HBr aq × × ×1000 mL 2 mol HBr 1 mol CaCO1 L HBr aq

3 3

3

= 1.52 g CaCO3

As long as we’re talking about acid-base chemistry, please name the following acids.

HBr(aq) hydrobromic acid

HBrO3(aq) bromic acid

HBrO(aq) hypobromous acid 5. (Fall 2012) On the “mini-exam” from earlier in semester, we considered the Earth-saving chemistry of

selenium, including some of the compounds listed below, and we introduced the selenate ion, 24SeO .

A. For each of the compounds shown, assign the oxidation numbers and write them in the spaces provided. SeO2 Se _______ O _______ K2Se K _______ Se _______ H2Se H _______ Se _______ H2SeO4 Se ______ H2SeO3 Se _______ B. In part A, the last compound is formed from the first compound according to the following chemical

equation: SeO2 + H2O H2SeO3

Is the above reaction a redox reaction? Explain your reasoning.

No. None of the oxidation numbers changed, so it is not a redox reaction.

C. The last three compounds in Part A form acids in water. Name the acids. (Recall that 24SeO

is selenate.)

H2Se(aq) ________________________________________ acid H2SeO4(aq) ________________________________________ acid H2SeO3(aq) ________________________________________ acid D. Which do you expect to be the stronger acid, H2SeO4(aq) or H2SeO3(aq)? Explain your reasoning.

H2SeO4(aq) should be the stronger acid.

Possible reasons: 1. The acidity of oxoacids increases with the oxidation state of the central atom.

2. H2SO4 is a stronger acid than H2SO3, so it seems reasonable that H2SeO4

would be a stronger acid than H2SeO3.


6. (Spring 2015) A reaction between sulfur and fluorine produced a gas that was believed to have the formula SFn, where n is an integer. The gas was collected, dried to remove water vapor, and weighted, giving a mass of 0.661 g. The gas occupied a volume of 150.3 mL at 23.1 °C and 751 mmHg.

A. Determine the molar mass of the gas.

150.3 mL => 0.1503 L 23.1 °C + 273.15 = 293.25 K 751 mmHg x (1 atm/760 mmHg) = 0.9981 atm

PV = nRT => n = PV/RT = [(0.9981 atm)( 0.1503 L)]/[(0.08206 L·atm/K·mol)(293.25 K)] = 0.006234 mol

0.661 g / 0.006234 mol = 106 g/mol B. What is the formula of the gas? In other words, what is the value of n?

molar mass of S + n(molar mass F) = 106 g/mol

32 g/mol + n(19 g/mol) = 106 g/mol

=> n = (106 - 32)/19 = 3.9 4 => SF4

C. Why did we have to dry the gas? (Be specific – don’t just say that water vapor is an impurity, so it had to

be removed. How does the presence of water vapor affect the experiment?)

The water vapor contributes to the total pressure, so the measured pressure (7.51 mmHg) is too high. Therefore the value calculated for n (# moles) exceeds the moles of nitrogen. It is the moles of nitrogen plus the moles of water vapor.

7. (Spring 2015) Appendicitis is caused by inflammation of the appendix, an organ of no apparent value (at least

not to modern humans) that we can live without just fine without. An appendectomy, the procedure in which the appendix is removed, can be done by conventional surgery, but the more common practice today involves laparoscopic surgery. The incision is smaller, so the recovery time is faster; however, the procedure requires using a scope, and limited visibility is an issue. But we know how to solve that problem.

In order to increase the work space and visibility, the abdominal cavity is inflated with a gas. The gases most commonly used for this procedure are CO2(g) and N2O(g). The pressures used range from around 10 to 25 mmHg gauge pressure; in other words, that is the pressure above atmospheric pressure. The World Laparoscopy Hospital website gives these values as typical:

Normal size human abdominal cavity needs 1.5 L CO2 to achieve intra-abdominal actual pressure of 12 mmHg [above atmospheric pressure].

A. Assuming a temperature of 37°C (normal body temperature) and an atmospheric pressure of 760 mmHg, how many moles of CO2 would be required for one laparoscopic appendectomy?

P = atmospheric pressure + 12 mmHg = 760 mmHg + 12 mmHg = 772 mmHg 772 mmHg x (1 atm/760 mmHg) = 1.016 atm T = 37C + 273.15 = 310 K PV = nRT => n = PV/RT = [(1.016 atm)(1.5 L)]/[(0.08206 L·atm/K·mol)(310 K)] = 0.060 mol

Suppose N2O gas was used instead of CO2. Would the number of moles required for the procedure increase, decrease, or remain the same? Explain your reasoning.

Remain the same. The nature of the gas does not affect PV = nRT, assuming ideal conditions.


B. The gases used in this procedure have advantages and disadvantages relative to each other. For example, N2O gas has a mild analgesic effect. On the other hand...

During prolonged laparoscopic procedure N2O should not be a preferred gas for pneumoperitoneum because it supports combustion better than air.

Consider this reaction with methane.

4N2O(g) + CH4(g) 4N2(g) + 2H2O(g) + CO2(g) ΔH = 1131 kJ/mol

1. How much heat would be released by reacting 25 g CH4 with an excess of N2O?

1 mol CH 1131 kJ released25 g CH × × = 1800 kJ16.04 g CH 1 mol CH

44

4 4

Question 1 (above)on Part B involves thermochemistry. We covered simple heat calculations on Friday, February 23, so this question is fair game for Exam 2, but instead of giving you ΔH = 1131 kJ/mol, I would tell you that this reaction releases 1131 kJ of energy. On Monday, February 26, we will identify enthalpy as a type of energy we conceptualize as heat. 2. Assign oxidation numbers for all of the atoms in this reaction. Write the oxidation numbers in

parentheses above the formula. ___ ___ ___ ___ ___ ___ ___ ___ ___ \ / \ / | \ / \ / 4N2O(g) + CH4(g) 4N2(g) + 2H2O(g) + CO2(g) Which element is oxidized? carbon Which is reduced? nitrogen (-4) to (+4) (+1) to (0) 8. (Fall 2015) We never got a chance to consider the “deflategate” controversy using the ideal gas law, so I

thought we might do it here.

Tom Brady and the New England Patriots were accused of deflating footballs used in their AFC championship game against the Indianapolis Colts on January 18, 2015. The NFL requires footballs to be inflated to a pressure between 12.5 and 13.5 “psi.” During halftime it was discovered that some of the footballs used by the Patriots were 2 psi below the minimum allowed pressure. Patriots supporters claimed that the drop in pressure was due to the difference in temperature between the locker room and the playing field.

A. Let’s assume that the footballs were filled to 12.5 psi before game and checked with a pressure gauge. That’s “gauge pressure” (psi-g) – the actual pressure (sometimes called “absolute” pressure) inside the ball would be 27.2 psi.

What’s the difference between gauge pressure and absolute pressure? In other words, explain why the numbers are different.

Absolute pressure is the pressure relative to zero, where a perfect vacuum is zero pressure. Gauge pressure is

the pressure relative to zero, where the pressure of the atmosphere is defined as zero pressure. B. According to reports, the temperature in the locker room before the game was 74°F (23°C) and the

temperature on the field at halftime was 48°F (9°C). Assuming a pressure of 27.2 psi at 23°C, what would the pressure be after the temperature dropped to 9°C?

Starting with PV = nRT, and moving constants to one side and variables to the other side gives

1 2 i f

1 2 i f

P P PP nR P = => = or =

T V T T T T

(+1) (-2) (-4) (+1) (0) (+1) (-2) (+4) (-2)


We have use absolute temperatures and pressures, so we have to change oC to K, but because the expression is set up as ratios, we can use psi without converting to atm. (But converting to atm is fine; you get the same answer.)

23 oC + 273.15 = 296 K 9 oC + 273.15 = 282 K

Rearranging this expression, the final pressure at halftime would be i f f

i

P T P =

T.

i f f

i

P T 27.2 psi x 282 K P = = = 25.9 psi

T 296 K

What is another assumption you make in the above calculation? Possible answers include: 1. volume remains constant 2. no air escaped football C. Would a temperature drop from 23°C to 9°C account for the footballs being 2 psi below the minimum

allowed pressure? Explain your reasoning. According to the calculations in Part B, the pressure should have dropped 27.2 psi – 25.9 psi = 1.3 psi. That is a

significant drop in pressure (5%), but not the 2 psi drop reported for some of the footballs. 9. (Fall 2015) Looks like we might finally be getting some fall weather, with cooler and less humid days.

Which is heavier, humid air or dry air?

Briefly explain your reasoning.

Dry air and humid air contains the same number of molecules of gas per volume (at the same temperature and pressure), but H2O is lighter than N2 or O2. Therefore, humid air is lighter (less dense) than dry air.

Humid air only “feels” heavier because the cooling caused by the evaporation of sweat is diminished in humid air. 10. (Fall 2015) We carried out the thermite reaction way back in Chapter 3 and used that reaction to talk about

limiting reagents and percent yield. Since then, it showed up in a recitation exercise and a homework problem. It’s a pretty impressive reaction. A large part of the driving force in the thermite reaction is based on the stability of aluminum oxide, as reflected in H of the following reaction.

2Al(s) + 3/2O2(g) Al2O3(s) H = 1669.8 kJ/mol

If 27 g of aluminum reacts with oxygen to form Al2O3(s), how much heat is released assuming the reaction goes to completion?

1 mol Al 1669.8 kJ released

27 g Al = 835.5 kJ26.98 g Al 2 mol Al

The abovequestion involves thermochemistry. We covered simple heat calculations on Friday, February 23, so this question is fair game for Exam 2, but instead of giving you ΔH = 1669.8 kJ/mol, I would tell you that this reaction releases 1669.8 kJ of energy. On Monday, February 26, we will identify enthalpy as a type of energy we conceptualize as heat.

Researchers at Purdue University have exploited this reactivity to develop a novel type of rocket fuel

consisting of aluminum nanoparticles dispersed in H2O(s). They nicknamed this propellant “ALICE” (get it? Al-ice) and noted that in addition to being environmentally friendly, ALICE could be produced on any extraterrestrial body that contains water. The recent discovery of water on Mars could draw new attention to this intriguing fuel. (International Journal of Aerospace Engineering, 2012)


times 3

flip, times 3

0 +1 -2 +3 -2 0

One can speculate that the reaction that causes ALICE to act as a propellant is

2Al(s) + 3H2O(s) Al2O3(s) + 3H2(g)

where the hydrogen gas is subsequently burned.

A. Given the molar enthalpies for the following reactions

2Al(s) + 3/2O2(g) Al2O3(s) H = 1669.8 kJ/mol H2(g) + 1/2O2(g) H2O(l) H = 285.8 kJ/mol H2O(s) H2O(l) H = 6.01 kJ/mol

determine H for the ALICE reaction

2Al(s) + 3H2O(s) Al2O3(s) + 3H2(g) 2Al(s) + 3/2O2(g) --> Al2O3(s) H = 1669.8 kJ/mol

3H2O(s) --> 3H2O(l) H = 3(6.01 kJ/mol)

3H2O(l) --> 3H2(g) + 3/2O2(g) H = 3(285.8 kJ/mol) 2Al(s) + 3H2O(s) --> Al2O3(s) + 3H2(g)

H = 1669.8 kJ/mol 3(6.01 kJ/mol) [3(285.8 kJ/mol)] = 794.4 kJ/mol Is the H you calculated consistent with Al-ice serving as a fuel? ________________

Give two reasons to support your answer. 1. energy was released (H < 0) one reason 2. the magnitude of H is fairly large 3. burning the H2 produced is also exothermic

B. Consider again the following reactions from Part A:

1. 2Al(s) + 3/2O2(g) Al2O3(s) H = 1669.8 kJ/mol 2. H2(g) + 1/2O2(g) H2O(l) H = 285.8 kJ/mol 3. H2O(s) H2O(l) H = 6.01 kJ/mol

For which, if any, reactions does H correspond to a molar enthalpy of formation? #1 and #2

C. Write the oxidation numbers in the spaces provided. ___ ___ ___ ___ ___ ___ | \ / \ / | 2Al(s) + 3H2O(s) Al2O3(s) + 3H2(g) 11. (Fall 2015) We looked at the decomposition of hydrogen peroxide in class. The reaction is slow, but if you

add a catalyst, it proceeds rather quickly.

2H2O2(aq) 2H2O(l) + O2(g)

A. If a catalyst is added to 154 mL of 0.882 M H2O2(aq), how many moles of O2(g) are produced, assuming the reaction goes to completion?

2 2 22 2 2

2 2 2 2

0.882 mol H O 1 mol O1 L154 mL H O (aq) = 0.0679 mol O

1000 mL 1 L H O (aq) 2 mol H O

B. The oxygen gas from the above procedure was collected and dried to remove water vapor. The volume of

O2(g) was 726 mL at 23.9C and 735 mmHg. Determine the amount (moles) of O2 gas collected. 726 mL x (1 L / 1000 mL) = 0.726 L


735 mmHg x (1 atm / 760 mmHg) = 0.967 atm

23.9 oC + 273.15 = 297 K

2PV (0.967 atm)(0.726 L)

PV = nRT => n = = = 0.288 mol OL atmRT 0.08206 (297 K)mol K

C. What was the percent yield for the reaction?

2

2

0.0288 mol Oactual yieldpercent yield = x 100% = x 100% = 42.4%

theoretical yield 0.0679 mol O

D. Why was it necessary to remove the water vapor? Most of us answered something like “removed water vapor because we needed the oxygen sample to be pure,”

but we will looking for a specific reason why the water vapor needed to be removed for the calculation to give the correct answer. If water vapor is not removed, then P = PO2 + PH2O, and using PV = nRT to solve for n would actually give moles of O2 plus moles of H2O(g). So we have to remove the water vapor before measuring pressure or we have to subtract the vapor pressure of water (which can be looked up as a function of temperature).

12. (Fall 2012) You may have heard that a few days ago Felix Baumgartner jumped from a balloon roughly 24

miles above Earth, breaking the record for the highest skydive. He reached speeds of more than 700 miles per hour and became the first skydiver to exceed the speed of sound during free fall. You will not be surprised to hear that this project was funded by Red Bull, but the underlying science is solid, and it is hoped that the work designing the suit worn by Baumgartner will lead to safer space suits for astronauts who must escape during a launch failure.

The helium-filled plastic balloon used to carry Baumgartner to the edge of space also required an impressive

bit of science and engineering. The thickness of the plastic is about one-tenth that of a Ziploc bag (more like a dry cleaner bag), yet it had to be strong enough to sustain very cold temperatures. The balloon was designed to expand to 3.0 107 cubic feet (8.5 108 L) to accommodate the extreme temperatures and pressures (67.8C and 0.027 mmHg, respectively) at the altitude required to break the record.

A. Determine the volume of the helium in the balloon just before it was released, assuming a pressure of 1.0

atm and a temperature of 23C. Rearranging PV = nRT to place the variables (P, V, T) on the left and the constants (R, n) on the right

gives

jump jump surface surface

jump surface

P V P VPV = nR = constant =T T T

Plugging in the known values

surface8760 mmHg1 atm V

atm(0.027 mmHg) (8.5 10 L)(-67.8 + 273.15)K (23 + 273.15)K

and solving for Vsurface gives

815

surface5

(0.027 mmHg)(296. K)(8.5 10 L)V = = 44,000 L

760 mmHg (205.3 K)


nitrogen

B. Using the conditions at the time of the jump (8.5 108 L, 67.8C, 0.027 mmHg), calculate the mass of helium in the balloon.

Rearranging PV = nRT gives

81 atm0.027 mmHg 8.5 10 L760 mmHg

1800 mol He (2 sf)L atm0.0821 (-67.8 + 273.15)KK mol

PVnRT

Helium is an inert gas, so it is monoatomic. Converting moles to grams…

4.003 g He1800 mol He 7200 g He1 mol He

C. If a pinhole was formed in the suit, which gas would escape faster, nitrogen or oxygen? _______________ Explain your reasoning. To receive full credit, your explanation should include the underlying scientific

principle as well as the property of the two gases that allowed you to answer the question.

The gases are at the same temperature in the suit, so they have the same kinetic energy, so the less massive gas will have the higher velocity.

13. (Fall 2012) The following reaction goes to completion in water: 2HCl(aq) + K2S(aq) H2S + 2KCl A. Is this reaction a precipitation reaction, an acid-base reaction, or a redox reaction?

acid-base reaction B. A solution was prepared by dissolving 0.716 g of K2S in 50 mL of water. It required 45.9 mL of HCl(aq)

to react completely with the K2S in solution. What is the concentration of the HCl solution in mol/L?

We can calculate the moles of HCl in 45.9 mL of solution from the mass of K2S that reacted.

22 9

2 2

1 mol K S 2 mol HCl0.716 g K S 0.0129 mol HCl110.26 g K S 1 mol K S

So the concentration in mol/L is

90.0129 mol HClmol HCl(aq)L solution 45.9 mL (1 L / 1000 mL)

0.283 M

C. It was observed during the course of the reaction that a gas was evolving from the solution. An inverted

tube filled with water was placed in the solution and a 4.11 mL sample of the gas was collected at 22.5 C and 128 mmHg. Set up the calculation that would give you the moles of gas collected. You do not have to carry out the calculation, but be sure to show the set up clearly.

Rearranging PV = nRT gives

5

1 atm 1 L128 mmHg 4.11 mL760 mmHg 1000 mL

2.85 10 mol gasL atm0.0821 (22.5 + 273.15 K)K mol

PVnRT


D. What else do you need to know to determine the molar mass of the gas?

We also need to know the mass of the gas collected. E. It is reasonable to expect that the gas formed was H2S(g), but when you calculated the molar mass of the

gas it was less than the molar mass of H2S. Why? A complete answer will explain why the calculated molar mass of the gas was different than the

molar mass of H2S, and also why the molar mass was less than the molar mass of H2S.

Given that the gas was collected over water, it will contain water vapor, so the total pressure gives the moles of H2S(g) plus the moles of H2O(g). We would need to subtract the partial pressure of water vapor at 22.5C to determine the moles of H2S(g).

Because the molar mass of H2O is less than the molar mass of H2S, the molar mass of the mixture will be less than the molar mass of H2S.

14. (Fall 2008) The proper assignments of oxidation numbers (ON) are necessary for recognizing and balancing

redox reactions (e.g. questions IV and V on page 3), but they are also useful for understanding other aspects of chemical reactivity. For example, I mentioned in class that oxidation numbers allow one to deduce the relative strength of analogous acids.

A. Assign the requested oxidation numbers for the following pairs of acids:

1. HNO2 HNO3

ON of N ______ ON of N ______

2. H2SO4 H2SO3

ON of S ______ ON of S ______

3. HClO4 HClO2

ON of Cl ______ ON of Cl ______

B. Circle the stronger acid in each of the above pairs.

C. What is the general relationship between oxidation number and acid strength?

D. On October 12, 2008, an oleum spill at a chemical plant in western Pennsylvania forced about 2,500 residents to evaluate. Oleum – also known as fuming sulfuric acid, H2SO4xSO3 – is especially dangerous because the sulfur trioxide escapes as a gas and reacts with moisture in the air to form sulfuric acid according to the reaction SO3(g) + H2O(g) H2SO4(g).

Is this a redox reaction? Explaining your reasoning. (*redox rxn, we’ll cover these after dot structures) 15. (Fall 2008) In class we talked about ways to determine the amount of iron in the solid mass formed when we

carried out the thermite reaction. We speculated that the mass could be reacted in a strong acid to get the iron into solution as Fe2+(aq), and that the remaining solid could be isolated, dried, and weighted to determine by difference the amount of iron in the mass.

Write a balanced chemical equation for the reaction of iron metal with HNO3(aq). (*redox rxn, we’ll cover these after dot structures)

A redox titration provides a more elegant way to determine the amount of iron in the solution. An aqueous solution of potassium permanganate reacts with Fe(NO3)2 according to the following chemical equation:

KMnO4(aq) + 8HNO3(aq) + 5Fe(NO3)2(aq) Mn(NO3)2(aq) + 5Fe(NO3)3(aq) + KNO3(aq) + 4H2O(l)


A solution of Fe(NO3)2(aq) was formed by reacting the product from a thermite reaction in HNO3(aq). That solution was titrated with a 0.537 M solution of KMnO4(aq) according to the above equation, and it took 121 mL to convert all of the Fe2+ to Fe3+.

How many moles of Fe2+ were contained in the original solution?

So, based on your above answer, how many grams of iron were formed in the thermite reaction?

16. (Fall 2008) Titration with an aqueous solution of potassium dichromate is a useful procedure for determining

the amount of iodide in solution. The unbalanced chemical equation for this reaction is given below. Assign the oxidation number of the atoms indicated below (all atoms except H and O).

___ ___ ___ ___ ___ ___ ___ ___ K2Cr2O7(aq) + HI(aq) CrI3(aq) + I2(s) + KI(aq) + H2O(l)

Identify the oxidizing agent and the reducing agent in the above reaction.

oxidizing agent ____________________ reducing agent ____________________

In order to use the above reaction to carry out a redox titration, you would need the balanced chemical equation, so please write the balanced equation below.

(*redox rxn, we’ll cover these after dot structures) 18. (Summer 2010) Several years ago an undergraduate working in my lab needed acetylene gas, HCCH(g), for an

experiment. Normally acetylene is purchased in a tank like other gases, but in the published experiment he was trying to repeat, the researchers generated acetylene by adding calcium carbide to a large volume of water and collecting the acetylene gas, according to this reaction

CaC2(s) + H2O(l) HCCH(g) + CaO(s)

Additional information for this problem is given in the info sheet passed out with this exam.

A. Is this a redox reaction? Explain your reasoning. (*redox rxn, we’ll cover these after dot structures)

B. If 0.035 moles of HCCH(g) is needed for the experiment, how many grams of CaC2 would be required?

C. Let’s assume you worked part B correctly and added that amount of CaC2(s) to water as described above, producing 718 mL of gas at 20.0C and 751 mm Hg. How many moles of HCCH(g) did you actually collect. Think about how you collected the gas as you consider your approach to solving this problem.

D. What was your percent yield in this reaction?


E. If the temperature of the gas collected increased, but the pressure remained the same, (circle one)

1. the volume of gas would increase remain the same decrease

2. the density of the gas would increase remain the same decrease

3. the speed of the gas molecules would increase remain the same decrease

4. the average kinetic energy of increase remain the same decrease the gas molecules would 19. (Fall 2010) A solid powder consisting of a mixture of Pb(NO3)2(s) and NaNO3(s) was discovered in a

warehouse. Authorities wish to evaluate the level of risk due to lead exposure, so a small amount of the solid was collected and submitted to the Department of Environmental Protection (DEP). A technician dissolved 0.629 g of the solid in water, and slowly added a 0.103 M solution of NaCl(aq), precipitating PbCl2(s) according to the reaction

Pb(NO3)2(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)

The solid precipitate was collected and dried, yielding 0.1676 g of PbCl2(s). The molar mass of PbCl2 is 278.106 g/mol.

A. What was the minimum volume (in mL) of 0.103 M NaCl(aq) required to precipitate all of the lead in the solution as PbCl2(s)?

B. What was the mass of Pb(NO3)2(s) present in the 0.629 g sample of solid tested by this procedure?

C. What is the mass percent Pb(NO3)2(s) in the sample?

D. Could we have used addition of NaCl(aq) to test for Hg(NO3)2(s) as described above? Explain briefly, i.e. resist the urge to fill the remaining space with your answer.

No. HgCl2 is soluble in water, so there would be no precipitate to isolate.


20. (Fall 2007) We carried out the following reaction in class:

2KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2KNO3(aq)

If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution, how many moles of Pb2+ were originally in the solution?

21. (Fall 2007) In class we used a simple apparatus consisting of a series of light bulbs of increasing wattage to

measure solution conductivity. The more light bulbs that light up, the greater the conductivity of the solution.

A. If we use this apparatus to test a 0.1 M solution of sulfuric acid, H2SO4(aq), all of the bulbs light up, but if we test a 0.1 M solution of sulfurous acid, H2SO3(aq), only a few bulbs light up. Explain this difference in behavior between the two solutions. Do not just say how the solutions are different – explain why this difference is reflected in the numbers of lit bulbs for the two solutions.

B. Write a balanced chemical equation for the reaction that would occur between the following solutions. As on the first page, be sure to include the correct states.

H2SO4(aq) + Ba(OH)2(aq) BaSO4(s) + 2H2O(l) Notice I did not specify whether I wanted the “molecular” equation or the net ionic equation. Why is that

not necessary in this reaction?

BaSO4 is an insoluble solid, and H2O is a very weak electrolyte => no ions to cancel

Strictly speaking the net ionic equation would be

H+(aq) + HSO42-(aq) + Ba2+(aq) + 2OH-(aq) BaSO4(s) + 2H2O(l)

The main point is that both the molecular and net ionic equation shows there are essentially no free ions in the products of the reaction.

C. If you start adding 0.1 M Ba(OH)2(aq) to 50mL of 0.1 M H2SO4(aq), an interesting thing happens: the number of bulbs that are lit starts to decrease until only one bulb is lit. If you keep adding Ba(OH)2(aq), the number of lit bulbs begins to increase until once again all of the bulbs are lit.

Why does the number of lit bulbs decrease as Ba(OH)2(aq) is added?

As Ba(OH)2(aq) is added, the ions from H2SO4(aq) – H+, HSO4-, and SO4

2- – are consumed to form BaSO4(s) and H2O(l) => conductivity of the solution decreases => fewer bulbs are lit

Is this consistent with your chemical equation in part B above? (It should be.) Explain.

Yes. Ions are consumed as BaSO4(s) and H2O(l) are formed.

D. At the point where the number of lit bulbs decreased to one, just before the number of light bulbs began to increase, what was the total volume of 0.1 M Ba(OH)2(aq) that had been added? Explain your reasoning. (You can explain in words or do a calculation.)


22. (Fall 2007) Commonly used gases in the laboratory are generally obtained from pressurized metal gas

cylinders, but for small amounts of occasionally used gases, it is sometimes easier just to prepare them chemically as needed. For example, nitrogen monoxide, NO(g), can be prepared in the lab by the following chemical reaction:

3Cu(s) + 8HNO3(aq) 2NO(g) + 3Cu(NO3)2(aq) + 4H2O(l)

A. One recipe calls for adding 0.50 L of 5.0 M HNO3(aq) to 25 g of copper metal.

1. How many moles of copper were initially present?

2. How many moles of HNO3 were added?

3. What is the limiting reagent? 4. How many moles of NO(g) could theoretically be produced in this manner?

B. When an earnest young CHM 1051L student followed this recipe, 5.8 L of gas was collected at a pressure of 746 torr and a temperature of 23.7C. Assuming all of the gas collected is nitrogen monoxide, how many moles of gas were actually collected?

Not asked, but fair game for Exam Two...

But these two questions are not fair game for Exam Two (Spring 2018) ...


23. (Fall 2007) A certain amount of gas is contained in a closed mercury manometer as shown below. What is the pressure of the gas in torr if h = 12.7 cm?

Assume the right hand tube is as long as you need it to be and no other parameters changed, would h increase, decrease, or remain the same if A. the moles of gas were doubled?

B. the moles of gas were kept the same, but the molecular weight of the gas was doubled?

C. the temperature was doubled?

D. the atmospheric pressure in the lab was doubled?

E. the mercury in the tube was replaced with water?

F. some gas was added to the vacuum at the top of the tube on the right?

G. the top of the tube on the right was broken off? 24. (Fall 2008) According to Wikipedia, “barium hydroxide is used in analytical chemistry for the titration of

weak acids, particularly organic acids.” Assuming Stephen Colbert has not undermined the integrity of this entry, we decided to use a barium hydroxide solution as the source of hydroxide to titrate a weak organic acid. Barium hydroxide is obtained as the octahydrate, Ba(OH)28H2O. What mass of Ba(OH)28H2O would be required to make 500.0 mL of a solution that is 0.1500 M in hydroxide ion?

25. (Fall 2008) In class we talked about ways to determine the amount of iron in the solid mass formed when we

carried out the thermite reaction. We speculated that the mass could be reacted in a strong acid to get the iron into solution as Fe2+(aq), and that the remaining solid could be isolated, dried, and weighted to determine by difference the amount of iron in the mass.

Write a balanced chemical equation for the reaction of iron metal with HNO3(aq). (*redox reaction)

A redox titration provides a more elegant way to determine the amount of iron in the solution. An aqueous solution of potassium permanganate reacts with Fe(NO3)2 according to the following chemical equation:

KMnO4(aq) + 8HNO3(aq) + 5Fe(NO3)2(aq) Mn(NO3)2(aq) + 5Fe(NO3)3(aq) + KNO3(aq) + 4H2O(l)

increase

increase

increase

decrease

no change

no change

decrease

= 11.82975 =>

11.83 g of Ba(OH)28H2O


A solution of Fe(NO3)2(aq) was formed by reacting the product from a thermite reaction in HNO3(aq). That solution was titrated with a 0.537 M solution of KMnO4(aq) according to the above equation, and it took 121 mL to convert all of the Fe2+ to Fe3+.

How many moles of Fe2+ were contained in the original solution?

So, based on your above answer, how many grams of iron were formed in the thermite reaction?

26. (Fall 2009) We normally consider “salty” and “sweet” to be totally different tastes associated with different

receptors on the tongue, but studies carried out with “trained testers” suggest that at low concentrations ( 0.01 M) table salt (sodium chloride, NaCl) actually tastes sweet. Reference: Food Qual. Pref. 2004, 15, 83.

A. How many grams of NaCl would you need to prepare 75 mL of a 0.25 M NaCl(aq) stock solution?

B. How much of the stock solution would you need to prepare 5 mL of 0.010 M NaCl(aq) need for the taste test?

27. (Fall 2009) A new route to nanocrystalline metal nitrides involves reacting metal chlorides with sodium azide

to form the metal azide and sodium chloride, followed by decomposition of the metal azide to give the metal nitride and nitrogen (Inorg. Chem. 2009, 48, 4470–4477). These reactions are shown below for the synthesis of manganese(II) nitride.

MnCl2(s) + 2 NaN3(s) Mn(N3)2(s) + 2 NaCl(s) (1)

Mn(N3)2(s) MnN(s) + 5/2 N2(g) (2) Note that reaction 1 is a solid state reaction, so ideally the reaction would be carried out with the

stoichiometrically correct amount of reactants (i.e. no limiting and excess reagents), giving the desired product in high yield with relatively little unreacted starting material as a contaminant.

(You may recall that we looked at this reaction in class, except we work a problem based on collecting the N2(g) and calculating the number of moles using PV=nRT.)

A typical reaction used 3.00 g of MnCl2 in this study.

A. How many grams of NaN3 would be required to give the correct stoichiometry based on reaction 1? (Molar masses are given in the handout.)


B. How many grams of Mn(N3)2 would be produced, assuming complete conversion of reactants to products.

C. In one early experiment using 3.00 g of MnCl2, only 2.3 g of Mn(N3)2 was obtained. Determine the percent yield for this experiment.

D. Even if the procedure was optimized to give 100% yield of Mn(N3)2, the desired compound may not be obtained as a pure compound. Explain and be specific with respect to this reaction.

The isolated solid could contain NaCl(s), the other product of the reaction. (Like the “iron” from our thermite reaction that also contained sand and/or Al2O3(s).

E. In reaction 2 is manganese oxidized or reduced? If so, which. Explain your reasoning. (*redox reaction) 28. (Fall 2009) Tripelennamine was developed and used as an antipruritic (anti-itch) and antihistamine, but it is

also a psychoactive drug and it has been largely replaced by newer antihistamines. Tripelennamine was initially synthesized by the legendary chemist Carl Djerassi, who is also a novelist and playwright (now you can feel good about the liberal studies credit you are getting for this course), but he is probably best known for his work on the first oral contraceptive (“the pill”). Djerassi received his first patent for tripelennamine shortly after getting his B.S. degree.

Djerassi would have certainly submitted a purified sample of his new compound for elemental analysis. Determine the empirical formula for tripelennamine given the results: 75.26% C; 8.29% H; 16.46% N.

Can you also determine the molecular formula from the above information? If so, determine the molecular

formula. If not, explain why you cannot determine the molecular formula from the information provided.


29. (Fall 2005) A convenient way to produce very high purity oxygen in the laboratory is to decompose

KMnO4(s) at high temperature according to the following equation

2KMnO4(s) K2MnO4(s) + MnO2(s) + O2(g)

If 2.50 L of O2(g) is needed at 1 atm and 20C, what mass of KMnO4(s) should be decomposed? You can assume the decomposition of KMnO4(s) goes to completion.

The oxygen gas was originally formed at “high temperature” before it cooled to 20C. Will the volume of the gas increase or decrease during the cooling process if kept a constant pressure? Explain your reasoning with either words or equations.

So as the temperature decreases, the volume of the gas will decrease. 30. (Fall 2009) It has been estimated that three trillion cubic feet (3 1012 ft3) of methane, CH4(g), is released into

the atmosphere every year (The New York Times, October 14, 2009). Not only would capturing that methane provide an additional source of energy, but it would also stop the leaking of a very potent greenhouse gas into the atmosphere (methane is 25 times more effective at trapping heat than an equal mass of carbon dioxide).

A. How many moles of methane are released into the atmosphere each year?

Methane as natural gas is generally measured in standard cubic feet at 60 F and 14.73 psia. To answer the above question you will need to convert some or all of the above units. I will take one step in that direction for you: 60 F = 16 C. The other information you need is given in the information handout; e.g. 1 ft3 = 28.32 L.


B. If the methane released into the atmosphere was measured at 25C, would the volume be less than, equal to, or greater than the volume at 16C?

_______________________________ Would the moles of methane released be less than, equal to, or greater than the moles of methane

calculated in part A? _______________________________ C. Given that the heat of combustion of methane is 890 kJ/mol, calculate the amount of energy that could be

obtained by the combustion of the methane that escapes each year?

(If you did not get an answer to part A, make up an answer, and I will grade part C accordingly.)

Part C of the previous question involves thermochemistry. We covered simple heat calculations on Friday, February 23, so this question is fair game for Exam 2. Problems 31-33 *are not* fair game for Exam 2.

greater than

equal to

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