Chapter 8: Problem 1
\(\quad \mathrm{CH}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\) Answer: Because oxygen appears as a pure elemental substance, we save it for last. Starting with carbon (an arbitrary choice), we note that there is one carbon on each side of the equation, meaning \(\mathrm{C}\) is balanced. There are four hydrogen atoms on the left and only two on the right. To fix this, we put a balancing coefficient \(2 \mathrm{in}\) front of the \(\mathrm{H}_{2} \mathrm{O}\) : \(\mathrm{CH}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\) This balances the hydrogen. Now it's time to balance the oxygen. There are two oxygen atoms on the left and four on the right (two from the \(\mathrm{CO}_{2}\) molecule and two from the two \(\mathrm{H}_{2} \mathrm{O}\) molecules). To balance the equation, we put a \(2 \mathrm{in}\) front of the \(\mathrm{O}_{2}\). \(\mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}\)
Short Answer
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