Gas Stoichiometry

Gas Stoichiometry

The ideal gas law can be used to determine the number of moles of a gas given pressure, volume and temperature. The number of moles of a gas in a chemical equation can be used to determine the number of moles of any other species in the same reaction. Hence, we can use the ideal gas law to help solve stoichiometry problems.

Let’s consider a sample problem.

Sample Problem

Calculate the volume of chlorine gas at STP that is required to completely react with 3.50 g of silver, using the following equation:

2Ag(s) + Cl2(g) —> 2AgCl(s)

Wanted: ? L Cl2(g) at STP

Given: 3.50 g Ag

To convert 3.50 g Ag to L Cl2(g) at STP, we need to first convert grams Ag to mole Ag, mol Ag to mol Cl2. Once we have mol Cl2 we can plug it into the Ideal Gas Law and solve for volume. There is an easier way, however. Since this gas is at STP, we can use the conversion factor that 1 mol of gas at STP has a volume of 22.414 L. Recall that 22.414 L is referred to as the molar volume.

Conversion Factors:

Converting 3.50 g Ag to mol Ag: 1 mol Ag = 107.9 g Ag

Converting mole Ag to mole Cl2: 2 mol Ag = 1 mol Cl2

Converting mole Cl2 to L Cl2 at STP: 1 mol gas at STP = 22.414 L

363L

 

3.50 g Ag at STP requires 0.363 L of chlorine gas.

 

Sample Problem

Calcium carbonate decomposes to form carbon dioxide and calcium oxide:

CaCO3(s) —> CO2(g) + CaO(s)

How many grams of calcium carbonate will be needed to form 3.45 liters of carbon dioxide at 740 mm Hg and 121 °C?

Wanted: ? g CaCO3

Given:

V = 3.45 L CO2

P = 740 mm Hg

T = 121 °C

Since this gas is not at STP, we will need to use the ideal gas law to solve for moles of CO2. Once we have moles of CO2, we can use stoichiometry to determine the grams of CaCO3 that will be needed.

The ideal gas law is:

(1) ideal-gas-law-small

 

Rewriting the equation to solve for moles:

(2) nPV

 

Because Rydberg’s constant is in the units L-atm-K-1-mol-1, we need to make sure our values for P, V and T are in the same units.

Temperature must be converted to Kelvin:

394K

 

And pressure must be converted to atmospheres:

97atm

 

Rewriting our Given:

Given:

V = 3.45 L CO2

P = 740 mm Hg = 0.97 atm

T = 121 °C = 394.15 K

Plugging these values into Equation (2):

(2) nPV

 

10mol

 

Now that we know 0.10 mol CO2 are produced, we can determine how many grams of CaCO3 are needed to produce it?

Our conversion factors are:

Converting 0.10 mol CO2 to mol CaCO3: Our balanced chemical equation tells us that 1 mol of CO2 is produced from 1 mol CaCO3.

Converting mol CaCO3 to g CaCO3 requires the molar mass of CaCO3: 100.09 g CaCO3 = 1 mol CaCO3

Setting up our dimensional analysis problem:

10gcaco3

 

10. g of CaCO3 are needed.

Sample Problem

When chlorine is added to acetylene, 1, 1, 2, 2-tetrachloroethane is formed:

2Cl2(g) + C2H2(g) —> C2H2Cl4(g)

How many liters of chlorine will be needed to make 75.0 grams of C2H2Cl4 at 1.90 atm and 30 °C?

Wanted: ? L Cl2

Given: 75.0 g C2H2Cl4

P = 1.90 atm

T = 30 °C

Since we are given grams of C2H2Cl4, we will use that to determine the number of moles of chorine gas required to produce it. Once we have moles chlorine gas, we can use the ideal gas law to determine volume.

To convert 75.0 g C2H2Cl4 to mol Cl2, we need the following conversion factors:

Converting 75.0 g C2H2Cl4 to mol C2H2Cl4 requires the molar mass of C2H2Cl4: 1 mol C2H2Cl4 = 167.836 g C2H2Cl4

Converting mol C2H2Cl4 to mol Cl2 requires the mole ratio between the two species, obtained from the balanced chemical reaction: 1 mol C2H2Cl4 = 2 mol Cl2

894mol

 

0.894 mol Cl2 will make 75.0 g of C2H2Cl4. We will plug this value and the above pressure and temperature values into the ideal gas law to compute the volume of Cl2 gas.

But first, we need to convert temperature to Kelvin:

303K

 

Given:

n= 0.894 mol Cl2

P = 1.90 atm

T = 30 °C = 303.15 K

The ideal gas law (1) is:

(1) ideal-gas-law-small

 

Rewriting to solve for V:

ideal-gas-law-V3

 

Plugging in our values for n, T and P:

117L

 

11.7 L Cl2 gas are needed.