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General Chemistry/Stoichiometry of Chemical Reactions/Calculations Involving Volumes of Gases

Stoichiometry of Chemical Reactions

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Calculations Involving Volumes of Gases

When products or reactants are gaseous in form at standard temperature and pressure, the molar volume of a gas and mole ratios can be used to convert the volume (in liters) of one substance to moles of another substance. One mole of an ideal gas has a volume of 22.4 L at standard temperature and pressure.

Calculations involving the volume of a gas are simpler than those involving a solid or a liquid because of a discovery made by the Italian chemist Amedeo Avogadro in 1811. Avogadro hypothesized that the volume of a gas depends only on the number of moles of the gas, regardless of the type of gas. Under equal conditions, the same number of moles of all gases have the same volume. Standard molar volume is the volume occupied by one mole of an ideal gas at STP (standard temperature and pressure), equal to 22.4 liters (L). The standard temperature is 273.15 kelvin (K), and the standard pressure is 1 atm.

The relationship among the moles of a gas, the volume of the gas in liters, and the molar volume is similar to the mass-to-mass formula.
Moles of gas=Volume of gasMolar volume{\text{Moles of gas}} = \frac{{{\text{Volume of gas}}}}{{{\text{Molar volume}}}}
Thus, the formula for the volume of gas is:
Volume of gas=(Moles of gas)(Molar volume)\text{Volume of gas}=(\text{Moles of gas})(\text{Molar volume})
Since molar volume is defined for one mole of a gas, in calculations, the conversion factor 22.4 liters per mole (L/mol) should be used.
A molar volume equation triangle is useful for remembering how to convert among volume, moles, and molar volume. The quantity in each section is equal to a function of the other two sections. The other two sections are multiplied if they are side by side, or divided if they are one on top of the other.
The steps involved are similar to the mass-to-mass method.

1. Write the balanced chemical equation for the reaction.
2. Determine the number of moles of the reactants.
3. Calculate the number of moles of the products.
4. Convert the moles of the products into volume.

Step-By-Step Example
Calculating the Volume of a Gas Produced by the Decomposition of Calcium Carbonate
Calculate the volume of carbon dioxide (CO2) produced at STP when 2.45 g of calcium carbonate (CaCO3) decomposes.
Step 1
Write the balanced equation.
CaCO3→CaO+CO2{\rm{CaCO}}_3\rightarrow\rm{CaO}+{\rm{CO}}_2
Step 2
Calculate the molar mass of CaCO3. Use the periodic table to determine the atomic mass of each atom. Then multiply the corresponding molar mass of each atom by the number of atoms shown in the formula.
Molar mass of CaCO3=(1)(40.08g/mol)+(1)(12.01g/mol)+(3)(16.00g/mol)=100.09g/mol\begin{aligned}{\text{Molar mass of }{\rm CaCO_3}}&=(1)(40.08{\rm{\; g/mol}})+(1)(12.01{\rm{\; g/mol}})+(3)(16.00{\rm{\; g/mol}})\\&= 100.09{\rm{\; g/mol}}\end{aligned}
Step 3
The number of moles of CaCO3 is the mass of CaCO3 divided by its molar mass.
Moles of CaCO3=Mass of CaCO3Molar mass of CaCO3=2.45g100.09g/mol=0.02448mol\begin{aligned}\text{Moles of }{\rm CaCO_3}&=\frac{\text{Mass of }{\rm CaCO_3}}{\text{Molar mass of }{\rm CaCO_3}}\\&=\frac{2.45\rm{\; g}}{100.09\rm{\; g/mol}}\\&=0.02448\rm{\; mol}\end{aligned}
Step 4
The balanced equation shows that the mole ratio of CaCO3 to CO2 is 1:1. Therefore, 0.02448 mol CO2 is produced.
Solution
Use the molar volume to convert the moles of gas to mass.
Volume of CO2=(Moles of CO2)(Molar volume)=(0.02448mol)(22.4L/mol)=0.548LCO2\begin{aligned}\text{Volume of }{\rm CO_2}&=(\text{Moles of }{\rm CO_2})(\text{Molar volume})\\&=(0.02448\rm{\; mol})(22.4\rm{\; L/mol})\\&=0.548\rm{\; L\; CO}_2\end{aligned}
Step-By-Step Example
Calculating the Volume of a Gas Produced by the Decomposition of Hydrogen Peroxide
Calculate the volume of oxygen gas produced when 5.00 g of hydrogen peroxide (H2O2) decomposes at STP.
Step 1
Write the balanced equation.
2H2O2→2H2O+O22{\rm H}_2{\rm O}_2\rightarrow2{\rm H}_2\rm O+{\rm O}_2
Step 2
Calculate the molar mass of hydrogen peroxide using the number of each type of atom in the formula and the molar mass of each element.
Molar mass of hydrogen peroxide=(2)(1.01g/mol)+(2)(16.00g/mol)=34.02g/mol\begin{aligned}\text{Molar mass of hydrogen peroxide}&=(2)(1.01\rm{\; g/mol})+(2)(16.00\rm{\; g/mol})\\&=34.02\rm{\; g/mol}\end{aligned}
Step 3
Convert the mass of hydrogen peroxide to moles using its molar mass.
Moles of hydrogen peroxide=Mass of hydrogen peroxideMolar mass of hydrogen peroxide=5.00g34.02g/mol=0.1470mol\begin{aligned}\text{Moles of hydrogen peroxide}&=\frac{\text{Mass of hydrogen peroxide}}{\text{Molar mass of hydrogen peroxide}}\\&=\frac{5.00\rm{\; g}}{34.02\rm{\; g/mol}}\\&=0.1470\rm{\; mol}\end{aligned}
Step 4
The balanced equation shows that the mole ratio of hydrogen peroxide to oxygen gas is 2:1. This means 2 mol hydrogen peroxide decomposes to produce 2 mol water and 1 mol oxygen gas.
Step 5
Use dimensional analysis to convert moles of hydrogen peroxide to moles of oxygen gas using the mole ratio.
Moles of O2=(0.1470molH2O2)(1molO22molH2O2)=0.0735molO2\begin{aligned}\text{Moles of }{\rm O_2}&=(0.1470\rm{\; mol\; H}_2\rm{O}_2)\!\left(\frac{1\rm{\; mol\; O}_2}{2\rm{\; mol\; H}_2\rm{O}_2}\right)\\&=0.0735\rm{\; mol\; O}_2\end{aligned}
Solution
Use the molar volume to convert the moles of oxygen gas to the volume of oxygen gas.
Volume of O2=(Moles of O2)(Molar volume)=(0.0735molO2)(22.4L/mol)=1.65L\begin{aligned}\text{Volume of }{\rm O_2}&=(\text{Moles of }{\rm O_2})(\text{Molar volume})\\&=(0.0735\rm{\; mol\; O}_2)(22.4\rm{\; L/mol})\\&=1.65\rm{\; L}\end{aligned}

Equations involving gases may be in volumes on both sides. In such cases, the steps to be followed are:

1. Write the balanced chemical equation.
2. Use the molar mass of a gas to convert moles to volume in liters.
3. Use the mole ratio to convert the number of moles of the products.
4. Use the molar volume to convert moles of the product to liters.

Step-By-Step Example
Calculating the Volume of Oxygen Required in a Combustion Reaction
How many liters of oxygen gas (O2) are required to burn 125 cm3 of ethane gas (C2H6) at STP?
Step 1
Write the balanced equation.
2C2H6+7O2→4CO2+6H2O2{\rm C}_2{\rm H}_6+7{\rm O}_2\rightarrow\;4{\rm{CO}}_2+6{\rm H}_2\rm O
Step 2
Convert the mass of C2H6 to volume in liters. Note that 1 cm3 is equal to 1 mL, and 1,000 mL is equal to 1 L.
volume of C2H6 in liters=(125cm3)(1mL1cm3)(1L1,000mL)=0.125L\begin{aligned}\text{volume of }{\rm C_{2}H_{6}}\text{ in liters}&=(125{\rm{\; cm}^3)}\!\left(\frac{1\rm{\; mL}}{1\rm{\; cm}^3}\right)\!\left(\frac{1\rm{\; L}}{1\rm{,}000{\; mL}}\right)\\&=0.125\rm{\; L}\end{aligned}
Step 3
Use the molar mass to convert liters of C2H6 to moles of C2H6.
Moles of C2H6=0.125L22.4L/mol=0.00558molC2H6\begin{aligned}\text{Moles of }{\rm C_{2}H_{6}}&=\frac{0.125\rm{\; L}}{22.4\rm{\; L/mol}}\\&=0.00558\rm{\; mol\; C}_2\rm{H}_6\end{aligned}
Step 4
The balanced equation shows that the mole ratio of C2H6 to O2 is 2:7. Use dimensional analysis to convert moles of C2H6 to moles of O2 using the mole ratio.
Moles of O2=(0.00558molC2H6)(7molO22molC2H6)=0.01953molO2\begin{aligned}\text{Moles of }{\rm O_{2}}&=(0.00558\rm{\; mol\; C}_2\rm{H}_6)\left(\frac{7\rm{\; mol \; O}_2}{2\rm{\; mol \; C}_2\rm{H}_6}\right)\\&=0.01953\rm{\; mol \; O}_2\end{aligned}
Step 5
Use the molar volume to find the volume of O2 in liters.
Volume of O2 in liters=(0.01953molO2)(22.4L/mol)=0.4375L\begin{aligned}\text{Volume of }{\rm O_{2}}\text{ in liters}&=(0.01953\rm{\; mol \; O}_2)(22.4\rm{\; L/mol})\\&=0.4375\rm{\; L}\end{aligned}
Solution
Use conversion factors to change liters to cubic centimeters, expressing the answer to three significant figures.
Volume of O2 in cubic centimeters=(0.4375L)(1,000mL1L)(1cm31mL)=437cm3\begin{aligned}\text{Volume of }{\rm O_{2}}\text{ in cubic centimeters}&=(0.4375\rm{\; L})\!\left(\frac{1\rm{,}000{\; mL}}{1\rm{\; L}}\right)\!\left(\frac{1\rm{\; cm}^3}{1\rm{\; mL}}\right)\\&=437\rm{\; cm}^3\end{aligned}
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