**Problem #1:** How many liters of O_{2} gas measured at 782.0 mmHg at 25.0 °C are required to completely react with 2.40 mol Al?

4 Al + 3 O_{2} ---> 2 Al_{2}O_{3}

**Solution:**

1) Determine the moles of O_{2} required to react with 2.40 mol of Al:

the molar ratio for Al and O_{2}is 4 to 3, so we set up the following ratio and proportion:

x = 1.80 moles of O

_{2}required

2) Determine volume of O_{2} at stated P and T:

PV = nRT ⇒ V = nRT / PComment: I converted 782.0 mmHg to atm.x = [ (1.80 mol) (0.08206) (298 K) ] / 1.029 atm

x = 42.8 L

**Problem #2:** What volume of carbon dioxide, at 1.00 atm and 112.0 °C, will be produced when 80.0 grams of methane (CH_{4}) is burned?

Video: the solution to the above problem

If we wished to determine the volume of O

_{2}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, C_{3}H_{8} 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:

C_{3}H_{8}+ 5 O_{2}---> 3 CO_{2}+ 4 H_{2}O

2) Determine moles of water vapor produced:

the molar ratio between propane used and water vapor produced is 1 to 4therefore, water vapor will be produced in the following ratio and proportion:

1 is to 4 as 1.50 is to xx = 6.00 moles of water vapor produced

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

PV = nRT ⇒ V = nRT / Px = [ (6.00 mol) (0.08206 L atm mol

^{-1}K^{-1}) (546 K) ] / 1.00 atmx = 269 L (to 3 sig. fig.)

**Problem #4:** Oxygen gas is sometimes prepared in labs by the thermal decomposition of potassium chlorate (KClO_{3}). The balanced chemical equation is as follows:

2 KClO_{3}(s) ---> 2 KCl(s) + 3 O_{2}(g)

If 5.150 grams decompose, what volume of O_{2} would be obtained at STP?

**Solution:**
1) Determine moles of KClO_{3}:

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

2) Determine moles of O_{2} produced:

the molar ratio of KClO_{3}used to O_{2}produced is 2 to 3therefore, oxygen will be produced in the following ratio and proportion:

2 is to 3 as 0.042006525 is to xx = 0.063009788 moles of O

_{2}produced

3) Determine the volume of O_{2} produced at STP:

PV = nRT ⇒ V = nRT / Px = [ (0.063009788 mol) (0.08206 L atm mol

^{-1}K^{-1}) (273 K) ] / 1.000 atmx = 1.412 L (to 4 sig. fig.)

**Problem #5:** When heated to high temperatures, silver oxide (Ag_{2}O) decomposes to form solid silver and oxygen gas. Calculate the volume of oxygen produced at STP by the decomposition of 7.44 g of Ag_{2}O.