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## 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?

## 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.