Gas Stoichiometry

1
Gas Stoichiometry

• A balanced equation shows the ratio of moles
  being used and produced
• Because of Avogrados principle, it also shows
  the ratio of volumes being used and produced

2
Example

• 2C4H10(g) 13O2(g) ? 8CO2(g) 10 H2O(g)
• Shows molar and volume ratios
• Questions
• How many moles of oxygen are required to produce
  35 moles of water?
• How many liters of butane does it take to
  completely react with 50 liters of oxygen?

3
Try It

• What volume of oxygen (O2) is needed to react
  with solid sulfur to form 3.5 L of SO2 gas?
• O2(g) S(s) ? SO2 (g)
• so 3.5 L O2 yields 3.5L SO2
• What volume of oxygen is needed to completely
  combust 2.36 L of methane gas?
• CH4(g) 2O2 (g) ? CO2(g) 2H2O(g)
• 2.36 L CH4 X 2 mol O2/ 1 mol CH4 4.72 L O2

4
Volume and Mass Problems

• A balanced equation indicates both moles and
  volume (for gases) and can convert either to mass
• Example
• N2(g) 3H2(g) ? 2NH3(g)
• If 5.00 L of nitrogen reacts completely at a
  constant pressure and temperature of 3.00 atm and
  298 K, how many grams of ammonia are produced?

5
Try It

• Given NH4NO3(s) ? N20(g) 2H2O(g)
• Calculate the mass of solid ammonium nitrate
  that must be used to obtain .100 L of dinitrogen
  oxide gas at a pressure of 3.00 atm and a
  temperature of 15o C.
• Hints change .100 L of N20 into moles
• calculate moles of NH4NO3 that reacted
• calculate the mass of the NH4NO3

6
Answers

• NH4NO3(s) ? N20(g) 2H2O(g)
• PV nRT or n PV / RT
• n 3.00 atm x .100 L / 288 K x .0821
  atm.L/K.mol
• .0127 mol N2O ? .0127 mol NH4NO3
• .0127 mol NH4NO3 x 79.035 g/mol
• 1.00 g NH4NO3

7
Try It

• 2. Solid potassium metal will react with Cl2 gas
  to form ionic potassium chloride. How many liters
  of Cl2 gas are needed to completely react with
  .204 g of potassium at STP?
• 2K(s) Cl2(g) ? 2KCl(aq)
• Steps convert grams of K into moles
• determine moles of Cl2 needed
• convert moles of Cl2 to liters

8
Answers

• 2K(s) Cl2(g) ? 2KCl(s)
• .204 g K x 1 mol / 39.098 g .00522 mol K
• .00522mol K x 1 mol Cl2 / 2 mol K .00261mol
  Cl2
• PV nRT so V nRT/P
• .00261 mol Cl2 x .0821atm.L/K.mol x 273 K /
  1.00 atm
• .0585 L or 5.85 x 10-2 L or 58.5 ml