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Chapter 3

Q. 3.12

In 2014 electric power plants in the United States consumed about 1.57 ×10^{11}  kg of natural gas. Natural gas is mostly methane, CH_{4}, so we can approximate the combustion reaction generating the energy by the equation:

CH_{4}(g) + 2 O_{2}(g) → CO_{2}(g) + 2 H_{2}O(g)

How many kilograms of CO_{2} were released into the atmosphere from these power plants in 2014?

Step-by-Step

Verified Solution

Collect, Organize, and Analyze We are asked to calculate the mass of CO_{2} produced from combustion of 1.57 ×10^{11}  kg of CH_{4}. The balanced equation tells us that 1 mole of carbon dioxide is produced for every 1 mole of methane consumed. We can change the mass of CH_{4} in kilograms to grams and then convert into moles of CH_{4} by multiplying by the molar mass of CH_{4}. In this reaction the moles of CH_{4} consumed is equal to the number of moles of CO_{2} produced, so the mole ratio is 1:1. Finally we convert moles of CO_{2} into kilograms of CO_{2}. We can combine all of these steps into a single calculation:

Solve The chemical equation is balanced as written (you should always check to be sure). To convert between grams and moles of CH_{4} and CO_{2}, we need to calculate the molar masses of these compounds. The molar mass of CH_{4} is

12.01  g/mol + 4(1.008  g/mol) = 16.04  g/mol  CH_{4}

and the molar mass of CO_{2} is

12.01  g/mol + 2(16.00  g/mol) = 44.01  g/mol  CO_{2}

Convert the mass of CH_{4} into moles:

1.57 \times 10^{11} \sout{ kg  CH_{4}} \times \frac{10^{3}  \sout{g}}{1  \sout{kg}} \times \frac{1  mol  CH_{4}}{16.04  \sout{g  CH_{4}}}=9.788 \times 10^{12}  mol  CH_{4}

Convert moles of CH_{4} into moles of CO_{2} by using the mole ratio from the balanced chemical equation:

9.788 \times 10^{12}  \sout{mol  CH_{4}} \times \frac{1  mol  CO_{2}}{1 \sout{ mol  CH_{4}}}=9.788 \times 10^{12}  mol  CO_{2}

Convert moles of CO_{2} into mass of CO_{2}:

9.788 \times 10^{12}  \sout{mol  CO_{2}} \times \frac{44.01  \sout{g}  CO_{2}}{1  \sout{mol  CO_{2}}} \times \frac{1  kg}{10^{3}  \sout{g}}=4.31 \times 10^{11}  kg  CO_{2}

We can combine the three separate calculations into a single calculation, and in subsequent problems we may not show the individual steps:

1.57 \times 10^{11}  \sout{kg  CH_{4}} \times \frac{10^{3}  \sout{g}}{1  \sout{kg}}\times \frac{1  \sout{mol  CH_{4}}}{16.04  \sout{g  CH_{4}}}\times \frac{1  \sout{mol  CO_{2}}}{1  \sout{mol  CH_{4}}}\times \frac{44.01  \sout{g}  CO_{2}}{1  \sout{mol  CO_{2}}}\times \frac{1  kg}{10^{3}  \sout{g}}=4.31 \times 10^{11}  kg  CO_{2}

Think About It The answer 4.31 \times 10^{11}  kg of CO_{2} is about 1.7% of the mass of CO_{2} we calculated for total annual CO_{2} production by combustion of fossil fuels.