# 10.7: Conversions Between Moles and Gas Volume

Small gas tanks are often used to supply gases for chemistry reactions. A gas gauge will give some information about how much is in the tank, but quantitative estimates are needed so the reaction will be able to proceed to completion. Knowing how to calculate needed parameters for gases is very helpful to avoid running out too early.

### Conversions Between Moles and Gas Volume

Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles. The equality of $$1 \: \text{mol} = 22.4 \: \text{L}$$ is the basis for the conversion factor.

Example 10.7.1

Many metals react with acids to produce hydrogen gas. A certain reaction produces $$86.5 \: \text{L}$$ of hydrogen gas at STP. How many moles of hydrogen were produced?

Solution:

Step 1: List the known quantities and plan the problem.

Known

• $$86.5 \: \text{L} \: \ce{H_2}$$
• $$1 \: \text{mol} = 22.4 \: \text{L}$$

Unknown

• moles of $$\ce{H_2}$$

Apply a conversion factor to convert from liters to moles.

Step 2: Calculate.

$86.5 \: \text{L} \: \ce{H_2} \times \frac{1 \: \text{mol} \: \ce{H_2}}{22.4 \: \text{L} \: \ce{H_2}} = 3.86 \: \text{mol} \: \ce{H_2}$

The volume of gas produced is nearly four times larger than the molar volume. The fact that the gas is hydrogen plays no role in the calculation.

Example 10.7.2

What volume does $$4.96 \: \text{mol}$$ of $$\ce{O_2}$$ occupy at STP?

Solution:

Step 1: List the known quantities and plan the problem.

Known

• $$4.96 \: \text{mol} \: \ce{O_2}$$
• $$1 \: \text{mol} = 22.4 \: \text{L}$$

Unknown

• Volume of $$\ce{O_2}$$

Step 2: Calculate.

$4.96 \: \text{mol} \times 22.4 \: \text{L/mol} = 111.1 \: \text{L}$

The volume seems correct given the number of moles.

Example 10.7.3

If we know the volume of a gas sample at STP, we can determine how much mass is present. Assume we have $$867 \: \text{L}$$ of $$\ce{N_2}$$ at STP. What is the mass of the nitrogen gas?

Solution:

Step 1: List the known quantities and plan the problem.

Known

• $$867 \: \text{L} \: \ce{N_2}$$
• $$1 \: \text{mol} = 22.4 \: \text{L}$$
• Molar mass $$\ce{N_2} = 28.02 \: \text{g/mol}$$

Unknown

• Mass $$\ce{N_2}$$

Step 2: Calculate.

We start by determining the number of moles of gas present. We know that 22.4 liters of a gas at STP equals one mole, so:

$867 \: \text{L} \times \frac{1 \: \text{mol}}{22.4 \: \text{L}} = 3.87 \: \text{mol}$

We also know the molecular weight of $$\ce{N_2}$$ $$\left( 28.0 \: \text{g/mol} \right)$$, so we can then calculate the weight of nitrogen gas in 867 liters:

$38.7 \: \text{mol} \times \frac{28 \: \text{g}}{1 \: \text{mol}} = 1083.6 \: \text{g} \: \ce{N_2}$