Created by Laura Peck, 2011Modern Chemistry

Example/Reminder Ex.1) Given the chemical equation:

Na2CO3(aq) + Ca(OH)2(s) 2NaOH(aq) + CaCO3(s) If we start with 2g Na2CO3 – how many grams NaOH are produced?

Grams start

1 molMolar mass(start)

(ratio)Mol finalMol start

Grams final

(molar mass)

2g Na2CO3 1 mol

105.967 g Na2CO3

2 mol

1 m0l

39.987 g NaOH

1 mol

2 x 1 x 2 x 39.987 / (105.965 x 1 x 1) = 1.509 g NaOH

Practice problems 1) Hydrogen and oxygen react under a

specific set of conditions to produce water according to the following: 2H2(g) + O2(g) 2H2O(g)

A) How many grams of hydrogen would be required to produce 5.0 g of water?

B) How many grams of oxygen would be required to produce 5.0 g of water?

5.0 g H2O

1 mol H2O

18.01 g H2O

2 mol

2 mol

2.014 g H2= 0.559 g H2

5.0 g H2O 1 mol H2O

18.01 g H2O

1 mol O22 mol H2O

31.98 g O2

1 mol O2 = 4.44 g O2

2) If 4.5 g of ethane, C2H6, undergo combustion according to: 2C2H6 + 5O2 4CO2 + 6H2O

A) how many grams of oxygen are required?

B) how many grams of carbon dioxide are produced?

C) how many grams of water are produced?

4.5 g C2H6 1 mol C2H6

30.062 g C2H6

5 mol O2

2 mol C2H61 mol O2

31.98 g O2= 11.967 g O2

4.5 g C2H6 x 1 mol C2H6 x 4 mol CO2 x 43.99g CO2 30.062 g C2H6 x 2mol C2H6 x 1 mol CO2= 13.17g CO2

4.5g C2H6 x 1 mol C2H6 x 6 mol H2O x 18.01g H2O 30.062g C2H6 x 2mol C2H6 x 1 mol H2O= 8.08g H2O

3) Sodium chloride is produced from its element

through a synthesis reaction. 2Na(s) + Cl2(g)

2NaCl(s)

A) What mass of each reactant would be required to produce 3.4g of sodium chloride?

B) What mass of each reactant would be required to produce 13.4g of sodium chloride?

3.4g NaCl x 1 mol NaCl x 2 mol Na x 22.989g Na 58.439g NaCl x 2mol NaCl x 1 mol Na= 1.34 g Na

3.4g NaCl x 1 mol NaCl x 1 mol Cl2 x 70.9g Cl2 58.439g NaCl x 2mol NaCl x 1 mol Cl2= 2.06g Cl

13.4 g NaCl x 1 mol NaCl x 2 mol Na x 22.98g Na 58.43 g NaCl x 2mol NaCl x 1 mol Na= 5.27g Na

13.4g NaCl x 1 mol NaCl x 1 mol Cl2 x 70.9 Cl2 58.43 g NaCl x 2 mol NaCl x 1 mol Cl2= 8.1 g Cl2

4) Iron is generally produced from iron ore through the following reaction in a blast furnace: 2Fe2O3(s) + 6CO(g) 4Fe(s) + 6CO2(g)

A) If 4.0 kg of Fe2O3 are available to react how many grams of CO are needed?

B) If 4.0 kg of Fe are produced, how many grams of Fe2O3 are needed?

4000 g Fe2O3 x 1 mol Fe2O3 x 6 mol CO x 28.009 g CO 159.687 g Fe2O3 x 2mol Fe2O3 x 1 mol CO

= 2104.8 g CO

4000 g Fe x 1 mol Fe x 2 mol Fe2O3 x 159.687 g Fe2O3 55.845 g Fe x 4 mol Fe x 1 mol Fe2O3

= 5718.93 g Fe2O3

5) Methanol, CH3OH, is an important industrial compound that is produced from the following reaction: CO(g) + 2H2(g)

CH3OH(g)

A) What mass of each reactant would be needed to produce 100.0 kg of methanol?

B) How many mols of H2 were consumed to generate 100.0 kg of methanol?

100000 g CH3OH x 1 mol CH3OH x 1 mol CO x 28.0 g CO 32.037 g CH3OH x 1 mol CH3OH x 1 mol CO= 87,398.94 g CO

100000 g CH3OH x 1 mol CH3OH x 2 mol H2 x 2.014 g H2 32.037 g CH3OH x 1 mol CH3OH x 1mol H2= 12,572.96 g H2

12572.96 g H2 x 1 mol H2 2.014 g H2= 6242.78 mol H2

5) Nitrogen combines with oxygen in the atmosphere during lightning flashes to form nitrogen monoxide, NO, which then reacts further with O2 to produce nitrogen dioxide, NO2 N2 + O2 2NO 2NO + O2 2NO2

A) What mass of NO2 is formed when NO reacts with 384 g of O2?

B) how many grams of NO are required to react with 455g of O2?

384 g O2 x 1 mol O2 x 2 mol NO2 x 46.065 g NO2 31.998 g O2 x 1 mol O2 x 1 mol NO2

= 1105.63 g NO2

455 g O2 x 1 mol O2 x 2 mol NO x 30.066 g NO 31.989 g O2 x 1 mol O2 x 1 mol NO

= 855.29 g NO

6) As early as 1938, the use of NaOH was suggested as a means of removing CO2 from the cabin of a spacecraft according to the following reaction: 2NaOH + CO2 Na2CO3 + H2O

A) If the average human body discharges 925.0 g of CO2 per day, how many grams of NaOH are needed each day for each person in the spacecraft?

B) How many mols of NaOH were needed for #A?

925.0 g CO2 x 1 mol CO2 x 2 mol NaOH x 39.995 g NaOH 44.008 g CO2 x 1 mol CO2 x 1 mol NaOH= 1681.3 g

NaOH

1681.3 g NaOH x 1 mol NaOH 39.995 g NaOH= 42.03 mol NaOH

7) The double-replacement reaction between silver nitrate and sodium bromide produces silver bromide, a component of photographic film. AgNO3 + NaBr AgBr + NaNO3

A) If 4.5 g of silver nitrate reacts, what mass of sodium bromide is required?

B) If 4.5 mol of silver nitrate reacts, how many mols of sodium bromide is required?

4.5 g AgNO3 x 1 mol AgNO3 x 1 mol NaBr x 102.88 g NaBr 169.932 g AgNO3 x 1mol AgNO3 x 1 mol NaBr= 2.72 g NaBr

4.5 mol AgNO3 x 1 mol NaBr or 1 mol NaBr = x 1 mol AgNO3 1mol AgNo3 4.5= 4.5 mol NaBr

8) In a soda-acid fire extinguisher, concentrated sulfuric acid reacts with sodium hydrogen carbonate to produce carbon dioxide, sodium sulfate and water H2SO4 + 2NaHCO2 2CO2 + Na2SO4 + 2H2O

A) How many grams of sodium hydrogen carbonate would be needed to react with 150.0 g of sulfuric acid?

B) how many moles of sodium sulfate would be produced from 150.0 g sulfuric acid?

150.0 g H2SO4 x 1 mol H2SO4 x 2 mol NaHCO2 x 68.004 g NaHCO2 98.07 g H2SO4 x 1 mol H2SO4 x 1 mol NaHCO2

= 208.02 g NaHCO2

150.0 g H2SO4 x 1 mol H2SO4 x 1 mol Na2SO4 98.07 g H2SO4 x 1 mol H2SO4

= 1.53 mol Na2SO4

Limiting reagent practice- find how many grams each reactant can produce- the one that produces the least is the limiting 9) Determine the limiting reagent when 2.o

mols of hydrochloric acid react with 2.5 mols of sodium hydroxide in a neutralization reaction. HCl + NaOH NaCl + H2O

2.0 mol HCl x 1 mol NaCl x 58.439 g NaCl 1 mol HCl x 1 mol NaCl

= 116.878 g NaCl

2.5 mol NaOH x 1 mol NaCl x 58.439 g NaCl 1 mol NaOH x 1 mol NaCl

= 146.09 g NaCl

10) If 2.50 mol of copper and 5.50 mol of silver nitrate are available to react by single replacement, identify the limiting reagent. Cu + AgNO3 Ag + CuNO3

2.5 mol Cu x 1 mol Ag x 107.86 g Ag 1 mol Cu x 1 mol Ag

= 269.65 g Ag

5.5 mol AgNO3 x 1 mol Ag x 107.86 g Ag 1 mol AgNO3 x 1 mol Ag= 593.23 g Ag

11) Sulfuric acid reacts with aluminum hydroxide by double replacement. H2SO4 + Al(OH)2 AlSO4 + 2H2O

A) If 30.0 g of sulfuric acid react with 25.0 g of aluminum hydroxide, identify the limiting reagent.

B) Determine the amount in moles of each product formed.

30.0 g H2SO4 x 1 mol H2SO4 x 2 mol H2O x 18.01 g H2O 98.075 g H2SO4 x 1 mol H2SO4 x 1 mol H2O= 11.02 g H2O

25.0 g Al(OH)2 x 1 mol Al(OH)2 x 2 mol H2O x 18.01 g H2O 60.992 g Al(OH)2 x 1 mol Al(OH)2 x 1 mol H2O

= 14.76 g H2O

11.02 g H2O x 1 mol H2O 18.01 g H2O

= 0.61 mols H2O

0.61 mols H2O x 1 mol AlSO4 or 1 mol AlSO4 = x 2 mol H2O 2 mol H2O 0.61

= 0.305 mol AlSO4