Multiple Choice
How many atoms of neon are present in 0.378 g of neon (Ne)?
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2. Atoms & Elements
Mole Concept
Multiple Choice
In the balanced chemical equation for the complete combustion of butane (C_4H_{10}), what is the mole ratio of butane (C_4H_{10}) to carbon dioxide (CO_2)?
A
1:2
B
4:8
C
1:4
D
2:8
Verified step by step guidance1
Write the unbalanced chemical equation for the complete combustion of butane: \(\mathrm{C_4H_{10} + O_2 \rightarrow CO_2 + H_2O}\).
Balance the carbon atoms first. Since butane has 4 carbon atoms, place a coefficient of 4 in front of \(\mathrm{CO_2}\): \(\mathrm{C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O}\).
Balance the hydrogen atoms next. Butane has 10 hydrogen atoms, so place a coefficient of 5 in front of \(\mathrm{H_2O}\) (because each water molecule has 2 hydrogens): \(\mathrm{C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O}\).
Balance the oxygen atoms last. On the right side, there are \(4 \times 2 = 8\) oxygen atoms from \(\mathrm{CO_2}\) and \(5 \times 1 = 5\) oxygen atoms from \(\mathrm{H_2O}\), totaling 13 oxygen atoms. Since oxygen gas is \(\mathrm{O_2}\), divide 13 by 2 to get the coefficient for \(\mathrm{O_2}\): \(\frac{13}{2}\).
To avoid fractional coefficients, multiply the entire equation by 2, resulting in the balanced equation: \(\mathrm{2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O}\). From this, the mole ratio of butane to carbon dioxide is \$2:8$.
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