Question

a major component of gasoline is octane ($ce{c_{8}h_{18}}$). when octane is burned in air, it chemically reacts with oxygen gas ($ce{o_{2}}$) to produce carbon dioxide ($ce{co_{2}}$) and water ($ce{h_{2}o}$).
what mass of water is produced by the reaction of 1.50 g of octane?
be sure your answer has the correct number of significant digits.
g

Explanation:

Step1: Balance the chemical equation

The reaction of octane (\(C_8H_{18}\)) with oxygen (\(O_2\)) to form carbon dioxide (\(CO_2\)) and water (\(H_2O\)) is a combustion reaction. The balanced equation is:

$$2C_8H_{18} + 25O_2 ightarrow 16CO_2 + 18H_2O$$

Step2: Calculate moles of octane

Molar mass of \(C_8H_{18}\): \(8\times12.01 + 18\times1.008 = 96.08 + 18.144 = 114.224\ \text{g/mol}\).
Moles of octane = \(\frac{\text{mass}}{\text{molar mass}} = \frac{1.50\ \text{g}}{114.224\ \text{g/mol}} \approx 0.01313\ \text{mol}\).

Step3: Relate moles of octane to water

From the balanced equation, \(2\ \text{mol}\ C_8H_{18}\) produces \(18\ \text{mol}\ H_2O\).
Moles of \(H_2O\) = \(0.01313\ \text{mol}\ C_8H_{18} \times \frac{18\ \text{mol}\ H_2O}{2\ \text{mol}\ C_8H_{18}} = 0.11817\ \text{mol}\).

Step4: Calculate mass of water

Molar mass of \(H_2O\): \(2\times1.008 + 16.00 = 18.016\ \text{g/mol}\).
Mass of \(H_2O\) = \(0.11817\ \text{mol} \times 18.016\ \text{g/mol} \approx 2.13\ \text{g}\).

Answer:

\(2.13\)