Questions: 0.390 mol of octane is allowed to react with 0.700 mol of oxygen. Which is the limiting reactant?

Solution Steps

The combustion of octane (\(\text{C}_8\text{H}_{18}\)) in oxygen (\(\text{O}_2\)) can be represented by the following balanced chemical equation:

\[ 2\text{C}_8\text{H}_{18} + 25\text{O}_2 \rightarrow 16\text{CO}_2 + 18\text{H}_2\text{O} \]

From the balanced equation, the mole ratio of octane to oxygen is:

\[ \frac{2 \text{ mol C}_8\text{H}_{18}}{25 \text{ mol O}_2} \]

To find the limiting reactant, we compare the mole ratio of the reactants provided:

\[ \text{Given: } 0.390 \text{ mol C}_8\text{H}_{18} \text{ and } 0.700 \text{ mol O}_2 \]

Calculate the required moles of \(\text{O}_2\) for 0.390 mol of \(\text{C}_8\text{H}_{18}\):

\[ 0.390 \text{ mol C}_8\text{H}_{18} \times \frac{25 \text{ mol O}_2}{2 \text{ mol C}_8\text{H}_{18}} = 4.875 \text{ mol O}_2 \]

Since only 0.700 mol of \(\text{O}_2\) is available, which is less than 4.875 mol, \(\text{O}_2\) is the limiting reactant.

\(\boxed{\text{O}_2 \text{ is the limiting reactant}}\)

Using the limiting reactant (\(\text{O}_2\)), we calculate the moles of water produced. From the balanced equation, the mole ratio of \(\text{O}_2\) to \(\text{H}_2\text{O}\) is:

\[ \frac{25 \text{ mol O}_2}{18 \text{ mol H}_2\text{O}} \]

Calculate the moles of \(\text{H}_2\text{O}\) produced from 0.700 mol of \(\text{O}_2\):

\[ 0.700 \text{ mol O}_2 \times \frac{18 \text{ mol H}_2\text{O}}{25 \text{ mol O}_2} = 0.5040 \text{ mol H}_2\text{O} \]

\(\boxed{0.5040 \text{ mol H}_2\text{O}}\)

Final Answer