Mass-Volume Stoichiometry Problems


Return to Stoichiometry menu

Problem #1: How many liters of O2 gas measured at 782.0 mmHg at 25.0 °C are required to completely react with 2.40 mol Al?
4 Al + 3 O2 ---> 2 Al2O3

Solution:

1) Determine the moles of O2 required to react with 2.40 mol of Al:

the molar ratio for Al and O2 is 4 to 3, so we set up the following ratio and proportion:

x = 1.80 moles of O2 required

2) Determine volume of O2 at stated P and T:

PV = nRT ⇒ V = nRT / P

x = [ (1.80 mol) (0.08206) (298 K) ] / 1.029 atm

x = 42.8 L

Comment: I converted 782.0 mmHg to atm.

Problem #2: What volume of carbon dioxide, at 1.00 atm and 112.0 °C, will be produced when 80.0 grams of methane (CH4) is burned?

Video: the solution to the above problem

If we wished to determine the volume of O2 required, we would use 9.9732 mol. This is because of the 1:2 molar ratio between methane and oxygen. To burn 4.9866 mol of methane requires 9.9732 mol of oxygen. Using PV = nRT, this is 315 L (I doubled the 157.5 figure from the video, not the 158.)


Problem #3: Propane, C3H8 reacts completely with oxygen to form carbon dioxide and water vapor. If 1.50 mole of propane is reacted with an excess of oxygen and the water vapor is collected and measured at 546 K and 1.00 atm, what volume of water vapor will be collected?

Solution:

1) Write a balanced chemical equation:

C3H8 + 5 O2 ---> 3 CO2 + 4 H2O

2) Determine moles of water vapor produced:

the molar ratio between propane used and water vapor produced is 1 to 4

therefore, water vapor will be produced in the following ratio and proportion:

1 is to 4 as 1.50 is to x

x = 6.00 moles of water vapor produced

3) Determine the volume of water vapor at the given temperature and pressure:

PV = nRT ⇒ V = nRT / P

x = [ (6.00 mol) (0.08206 L atm mol-1 K-1) (546 K) ] / 1.00 atm

x = 269 L (to 3 sig. fig.)


Problem #4: Oxygen gas is sometimes prepared in labs by the thermal decomposition of potassium chlorate (KClO3). The balanced chemical equation is as follows:

2 KClO3(s) ---> 2 KCl(s) + 3 O2(g)

If 5.150 grams decompose, what volume of O2 would be obtained at STP?

Solution: 1) Determine moles of KClO3:

5.150 g / 122.6 g mol-1 = 0.042006525 mol (I kept some guard digits.)

2) Determine moles of O2 produced:

the molar ratio of KClO3 used to O2 produced is 2 to 3

therefore, oxygen will be produced in the following ratio and proportion:

2 is to 3 as 0.042006525 is to x

x = 0.063009788 moles of O2 produced

3) Determine the volume of O2 produced at STP:

PV = nRT ⇒ V = nRT / P

x = [ (0.063009788 mol) (0.08206 L atm mol-1 K-1) (273 K) ] / 1.000 atm

x = 1.412 L (to 4 sig. fig.)


Problem #5: When heated to high temperatures, silver oxide (Ag2O) decomposes to form solid silver and oxygen gas. Calculate the volume of oxygen produced at STP by the decomposition of 7.44 g of Ag2O.

Video: the solution to the above problem