Basic Stoichiometry
description
Transcript of Basic Stoichiometry
![Page 1: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/1.jpg)
Basic Stoichiometry
Pisgah High SchoolM. Jones
Revision history: 5/16/03, 02/04/12, 04/27/12
Part One
![Page 2: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/2.jpg)
The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.
![Page 3: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/3.jpg)
Stoichiometry deals with the amounts of reactants and products in a chemical reaction.
![Page 4: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/4.jpg)
Stoichiometry deals with moles.
![Page 5: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/5.jpg)
Recall that …
1 mole = 22.4 L of any gas at STP
1 mole = the molar mass
molecules1 mole = 6.022 x 1023 atoms or
![Page 6: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/6.jpg)
The word mole is one that represents a very large number.
… “mole” means 6.022 x 1023
Much like “dozen” means 12,
![Page 7: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/7.jpg)
The key to doing stoichiometry is the balanced chemical equation.
2 H2 + O2 2 H2O22
![Page 8: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/8.jpg)
The coefficients give the relative number of atoms or molecules of each reactant or product …as well as the number
of moles.
2 H2 + O2 2 H2O
![Page 9: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/9.jpg)
2 H2 + O2 2 H2O2 moleculesof hydrogen
1 molecule of oxygen
2 moleculesof water
Two molecules of hydrogen combine with one molecule of oxygen to make two molecules of water.
![Page 10: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/10.jpg)
2 H2 + O2 2 H2O
The balanced chemical equation also gives the smallest integer number of moles.
2 moleculesof hydrogen
1 molecule of oxygen
2 moleculesof water
![Page 11: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/11.jpg)
2 H2 + O2 2 H2O
The balanced chemical equation also gives the smallest integer number of moles.
2 molesof hydrogen
1 moleof oxygen
2 molesof water
![Page 12: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/12.jpg)
2 H2 + O2 2 H2O2 molesof hydrogen
1 moleof oxygen
2 molesof water
Two moles of hydrogen combine with
one mole of oxygen to make two moles of water.
![Page 13: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/13.jpg)
How can we show that this is really true?
Consider the combustion of hydrogen in oxygen …
![Page 14: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/14.jpg)
2 H2 + O2 2 H2O2 molesof hydrogen
1 moleof oxygen
2 molesof water
What do each of the reactants and product weigh?
![Page 15: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/15.jpg)
2 H2 + O2 2 H2O2 molesof hydrogen
1 moleof oxygen
2 molesof water
2 x 2.0 g 1 x 32.0 g 2 x 18.0 g
4.0 g 32.0 g 36.0 g+ =
![Page 16: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/16.jpg)
2 H2 + O2 2 H2O2 molesof hydrogen
1 moleof oxygen
2 molesof water
4.0 g 32.0 g 36.0 g+ =
The Law of Conservation of Matter
![Page 17: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/17.jpg)
2 H2 + O2 2 H2O2 molesof hydrogen
1 moleof oxygen
2 molesof water
4.0 g 32.0 g 36.0 g+ =
Matter can neither be created nor destroyed, only changed in form.
![Page 18: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/18.jpg)
The oxidation of iron
![Page 19: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/19.jpg)
Consider the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2
4 moles 3 moles 2 moles
The coefficients give the ratio of moles.
![Page 20: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/20.jpg)
Consider the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2If these react … then we have the following:
8 moles 6 moles 4 moles2 moles 1.5 moles 1 mole0.50 mol 0.375 mol 0.25 mol
4 moles 3 moles 2 moles
![Page 21: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/21.jpg)
Consider the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2
Fe mole 4
O mole 3 Fe mole 5.0 2 0.375 mole O2
The 0.375 moles was not as easy to predict.
![Page 22: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/22.jpg)
Fe mole 4
O mole 3 Fe mole 5.0 2
Consider the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2
Use a conversion factor to determine the number of moles of an unknown.
0.375 mole O2
The 0.375 moles was not as easy to predict.
![Page 23: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/23.jpg)
Fe mole 4
O mole 3 Fe mole 5.0 2
Consider the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2
The conversion factor comes from the coefficients in the balanced equation.
0.375 mole O2
![Page 24: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/24.jpg)
Back to the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2
4 moles 3 moles 2 moles
Calculate the masses of these moles.
![Page 25: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/25.jpg)
Back to the oxidation of iron:
Fe(s) + O2(g) Fe2O3(s)4 3 2
4 moles 3 moles 2 moles4 x 55.8 g 3 x 32.0 g 2 x 159.6 g
319.2 g96.0 g223.2 g + =
Mass is conserved.
![Page 26: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/26.jpg)
The decomposition of ammonium carbonate
![Page 27: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/27.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
Suppose 96.0 grams of ammonium carbonate decomposes. How many grams of each of the gases will be produced?
![Page 28: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/28.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
96.0 grams ? g ? g ? g
The coefficients tell moles, not grams.
Convert 96.0 g (NH4)2CO3 to moles.
![Page 29: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/29.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
96.0 grams ? g ? g ? g
Find the molar mass of ammonium carbonate.
2x14 + 8 +12 + 3x16 = 96.0 g/mol
![Page 30: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/30.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
96.0 grams ? g ? g ? g
Isn’t that convenient! We have one mole of ammonium
carbonate.
![Page 31: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/31.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
96.0 grams ? g ? g ? g1 mol 1 mol1 mol2 mol
2 moles of ammonia are produced, along with 1 mole of carbon dioxide
and 1 mole of water vapor.
![Page 32: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/32.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
96.0 grams ? g ? g ? g
How many grams of each product are formed?
1 mol 1 mol1 mol2 mol
2 x 17.0g 44.0 g 18.0 g
![Page 33: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/33.jpg)
Now consider the decomposition of solid ammonium carbonate.
(NH4)2CO3 2 NH3 + CO2 + H2O
96.0 grams ? g ? g ? g
34.0 g + 44.0 g + 18.0 g =
1 mol 1 mol1 mol2 mol
2 x 17.0g 44.0 g 18.0 g
96.0 g
![Page 34: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/34.jpg)
The reaction between dinitrogen pentoxide and
water
![Page 35: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/35.jpg)
Consider the reaction between dinitrogen pentoxide and water.
What kind of reaction is it?Is it a …
Double replacement reaction?Combustion reaction?Decomposition reaction?Single replacement reaction?
So it must be a synthesis reaction
![Page 36: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/36.jpg)
Consider the reaction between dinitrogen pentoxide and water.
Which kind of synthesis reaction is it?
1. Hydrogen + nonmetal = binary acid2. Metal + nonmetal = salt3. Metal + water = base4. Nonmetal + water = ternary acid
N2O5 + HOH 2 HNO3
HNO3 is a ternary acid; HNO3 is nitric acid
![Page 37: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/37.jpg)
N2O5 + HOH 2 HNO3
Suppose we needed to make 100.0 grams of nitric acid.
How many grams of dinitrogen pentoxide would we need to
react with excess water?
![Page 38: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/38.jpg)
N2O5 + HOH 2 HNO3100.0 g??? g
3
33 HNO g 63.0
HNO mole HNO g 0.100
= 1.59 mole HNO3
1.59 mole0.794 mole
52
5252 ON mole 1
ON g .0108 ON mole 794.0 = 85.7 g N2O5
1.59 mol/2
![Page 39: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/39.jpg)
N2O5 + HOH 2 HNO3100.0 g85.7 g
1.59 mole0.794 mole1.59 mol/2
But. Suppose that we find out that there are only 60.0 grams of dinitrogen pentoxide.
How many grams of nitric acid could we actually make?
![Page 40: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/40.jpg)
N2O5 + HOH 2 HNO3??? g60.0 g
3
33 HNO mol 1.00
HNO g 0.63 HNO mol 11.1 = 70.0 g HNO3
1.11 mole0.556 mole
52
5252 ON g 108
ON mol 1 ON g 0.06 = 0.556 mol N2O5
0.556 mol x 2
![Page 41: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/41.jpg)
Description of stoichiometry problems
![Page 42: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/42.jpg)
Stoichiometry problems will usually take one of the following forms:
1. Mole-mole problem where you might be given moles and asked to find moles of another substance.
2. Mole-mass problem where you might be given moles and asked find the mass of another substance.
![Page 43: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/43.jpg)
3. Mass-mass problem where you might be given a mass and asked to find the mass of another substance.
4. Mass-volume problem where you might be given a mass and asked to find the volume of a gas.
5. Volume-volume problem where you might be given a volume and asked to find another volume.
![Page 44: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/44.jpg)
Volume-volume stoichiometry problems are easiest when you use Gay-Lussac’s Law.
“The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.”
![Page 45: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/45.jpg)
Simply put, Gay-Lussac’s Law says this:
The volumes of the gases are in the same ratio as the coefficients in the balanced equation.
2 H2 + O2 2 H2O2 moles 1 mole 2 moles2 L 1 L 2 L
![Page 46: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/46.jpg)
Applications of Gay-Lussac’s Law
![Page 47: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/47.jpg)
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
1 mol 5 mols 3 mols 4 mols
Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required?
17.5 L O2
1 L 5 L 3 L 4 L
HC L 1
O L 5 HC L .53
83
283
![Page 48: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/48.jpg)
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
1 mol 5 mols 3 mols 4 mols
Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O2
How many liters of carbon dioxide gas and water vapor at STP would be produced?
10.5 L CO2 and 14.0 L H2O
1 L 5 L 3 L 4 L
![Page 49: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/49.jpg)
CH4(g) + 4 Cl2(g) CCl4(g) + 4 HCl(g)
When methane gas is allowed to react with an excess of chlorine gas, tetrachloromethane and hydrogen chloride gas will be produced.
How many L of methane will react with 0.800 L of chlorine gas at STP?
Cl L 4
CH L 1 Cl L 800.0
2
42 0.200 L Cl2
![Page 50: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/50.jpg)
Stoichiometry problems involving gases
![Page 51: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/51.jpg)
Find the mass of HOCl that can be produced when 2.80 L of chlorine gas at STP reacts with excess hydrogen peroxide.
Cl2(g) + H2O2 (l) 2 HOCl (l)2.80 L ??? g
Convert 2.80 L of Cl2 gas at STP to moles
![Page 52: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/52.jpg)
Cl2(g) + H2O2 (l) 2 HOCl (l)2.80 L ??? g
Cl L 22.4
Cl mol 1 Cl L .802
2
22 0.125 mol Cl2
0.125 mol.125 x 2
0.250 mol
HOCl mol 1
HOCl g 52.5 HOCl mol 250.0 13.1g
HOCl
![Page 53: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/53.jpg)
The reaction between copper and nitric acid
![Page 54: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/54.jpg)
Will copper dissolve in acids?
Cu + 2HCl CuCl2 + 2H2 (g) No Reaction
Most metals react with HCl to produce a metal chloride solution and H2 gas.
Not copper
Consider hydrochloric acid
![Page 55: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/55.jpg)
Will copper dissolve in acids?
Cu + 2HCl CuCl2 + 2H2 (g) No Reaction
Consider hydrochloric acid
Copper is below hydrogen in the activity series. Copper metal will only replace elements that are below it in the activity series.
![Page 56: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/56.jpg)
What about other acids?
The same is true for all acids except nitric acid
Cu + HBr NRCu + HI NRCu + HF NRCu + H2SO4 NRCu + HC2H3O2 NR
![Page 57: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/57.jpg)
Nitric acid is the only acid that will dissolve copper.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
A beaker contains a penny and some nitric acid is added.
![Page 58: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/58.jpg)
Nitric acid is the only acid that will dissolve copper.
The penny begins to disappear and the solution turns blue-green and a brown gas is given off.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
![Page 59: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/59.jpg)
Nitric acid is the only acid that will dissolve copper.
The gas produced in the reaction is NO, which is colorless. Reddish brown NO2 forms when NO reacts with the oxygen in the air.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
![Page 60: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/60.jpg)
Nitric acid is the only acid that will dissolve copper.
The penny is gone and the solution turns a dark blue. The brown NO2 gas escapes from the open beaker.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
![Page 61: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/61.jpg)
Nitric acid is the only acid that will dissolve copper.
Calculate the volume of NO gas at STP when 20.0 grams of copper dissolves.
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
![Page 62: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/62.jpg)
Nitric acid is the only acid that will dissolve copper.
20.0 g ??? L
Cu g 63.5
Cu mol 1Cu g 0.20 0.315 mol Cu
0.315 mol 0.210 mol x 0.667
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
![Page 63: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/63.jpg)
Nitric acid is the only acid that will dissolve copper.
20.0 g
NO mol 1
NO L 22.4 NO mol 0.210 4.70 L NO
0.315 mol
??? L
0.210 mol
3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O
x 0.667
![Page 64: Basic Stoichiometry](https://reader036.fdocuments.in/reader036/viewer/2022070406/56814274550346895dae9b8b/html5/thumbnails/64.jpg)
Part Two of Basic Stoichiometry
will include the gas laws.