# 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) + Cl_{2}(g) —> 2AgCl(s)

Wanted: ? L Cl_{2}(g) at STP

Given: 3.50 g Ag

To convert 3.50 g Ag to L Cl_{2}(g) at STP, we need to first convert grams Ag to mole Ag, mol Ag to mol Cl_{2}. Once we have mol Cl_{2} 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 Cl_{2}: 2 mol Ag = 1 mol Cl_{2}

Converting mole Cl_{2} to L Cl_{2} at STP: 1 mol gas at STP = 22.414 L

*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:

CaCO_{3}(s) —> CO_{2}(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 CaCO_{3}

Given:

V = 3.45 L CO_{2}

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 CO_{2}. Once we have moles of CO_{2}, we can use stoichiometry to determine the grams of CaCO_{3} that will be needed.

The ideal gas law is:

Rewriting the equation to solve for moles:

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:

And pressure must be converted to atmospheres:

Rewriting our Given:

Given:

V = 3.45 L CO_{2}

P = 740 mm Hg = 0.97 atm

T = 121 °C = 394.15 K

Plugging these values into Equation (2):

Now that we know 0.10 mol CO_{2} are produced, we can determine how many grams of CaCO_{3} are needed to produce it?

Our conversion factors are:

Converting 0.10 mol CO_{2} to mol CaCO_{3}: Our balanced chemical equation tells us that 1 mol of CO_{2} is produced from 1 mol CaCO_{3}.

Converting mol CaCO_{3} to g CaCO_{3} requires the molar mass of CaCO_{3}: 100.09 g CaCO_{3} = 1 mol CaCO_{3}

Setting up our dimensional analysis problem:

10. g of CaCO_{3} are needed.

Sample Problem

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

2Cl_{2}(g) + C_{2}H_{2}(g) —> C_{2}H_{2}Cl_{4}(g)

How many liters of chlorine will be needed to make 75.0 grams of C_{2}H_{2}Cl_{4} at 1.90 atm and 30 °C?

Wanted: ? L Cl_{2}

Given: 75.0 g C_{2}H_{2}Cl_{4}

P = 1.90 atm

T = 30 °C

Since we are given grams of C_{2}H_{2}Cl_{4}, 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 C_{2}H_{2}Cl_{4} to mol Cl_{2}, we need the following conversion factors:

Converting 75.0 g C_{2}H_{2}Cl_{4} to mol C_{2}H_{2}Cl_{4} requires the molar mass of C_{2}H_{2}Cl_{4}: 1 mol C_{2}H_{2}Cl_{4} = 167.836 g C_{2}H_{2}Cl_{4}

Converting mol C_{2}H_{2}Cl_{4} to mol Cl_{2} requires the mole ratio between the two species, obtained from the balanced chemical reaction: 1 mol C_{2}H_{2}Cl_{4} = 2 mol Cl_{2}

0.894 mol Cl_{2} will make 75.0 g of C_{2}H_{2}Cl_{4}. We will plug this value and the above pressure and temperature values into the ideal gas law to compute the volume of Cl_{2} gas.

But first, we need to convert temperature to Kelvin:

Given:

*n*= 0.894 mol Cl_{2}

P = 1.90 atm

T = 30 °C = 303.15 K

The ideal gas law (1) is:

Rewriting to solve for *V*:

Plugging in our values for *n*, *T* and *P*:

*11.7 L Cl _{2} gas are needed.*