Gas Stoichiometry
• A balanced equation shows the ratio of
moles being used and produced
• Because of Avogrado’s principle, it also
shows the ratio of volumes being used and
produced
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?
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
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?
Try It
1. 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
Answers
1. 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 NH NO x 79.035 g/mol
= 1.00 g NH4NO3
4
3
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
Answers
2K(s) + Cl2(g)  2KCl(s)
.204 g K x 1 mol / 39.098 g = .00522 mol K
.00522mol K x 1 mol Cl / 2 mol K = .00261mol Cl2
2
PV = nRT so V = nRT/P
= .00261 mol Cl x .0821atm L/K mol x 273 K / 1.00 atm
2
.
.
= .0585 L or 5.85 x 10-2 L or 58.5 ml