Question

a major component of gasoline is octane (c₈h₁₈). when liquid octane is burned in air it reacts with oxygen (o₂) gas to produce carbon dioxide gas and water vapor. calculate the moles of carbon dioxide produced by the reaction of 0.070 mol of oxygen. be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

Explanation:

Step1: Write the balanced chemical equation

The balanced equation for the combustion of octane is $2C_8H_{18}(l)+25O_2(g)
ightarrow16CO_2(g) + 18H_2O(g)$.

Step2: Determine the mole - ratio

From the balanced equation, the mole - ratio of $O_2$ to $CO_2$ is $25:16$.

Step3: Calculate moles of $CO_2$

Let $n_{CO_2}$ be the moles of $CO_2$ and $n_{O_2}=0.070$ mol. Using the mole - ratio, we have the proportion $\frac{n_{CO_2}}{n_{O_2}}=\frac{16}{25}$. So $n_{CO_2}=n_{O_2}\times\frac{16}{25}$. Substituting $n_{O_2} = 0.070$ mol, we get $n_{CO_2}=0.070\times\frac{16}{25}=0.0448$ mol. Rounding to two significant digits (since 0.070 has two significant digits), $n_{CO_2}=0.045$ mol.

Answer:

$0.045$ mol