Question

$$2\\ce{c8h18} + 25\\ce{o2} \ ightarrow 16\\ce{co2} + 18\\ce{h2o}$$

in a combustion engine,
35.0 g of octane, \ce{c8h18}, reacts with
56.0 l of oxygen at stp.
what is the theoretical yield, in moles, of
carbon dioxide produced?
? mol \ce{co2}
be careful to check for the limiting reactant.

Explanation:

Step1: Calculate moles of octane

Molar mass of \( C_8H_{18} \) is \( 8\times12.01 + 18\times1.008 = 114.23 \, g/mol \).
Moles of \( C_8H_{18} = \frac{35.0 \, g}{114.23 \, g/mol} \approx 0.3064 \, mol \).

Step2: Calculate moles of \( O_2 \) at STP

At STP, 1 mol gas = 22.4 L.
Moles of \( O_2 = \frac{56.0 \, L}{22.4 \, L/mol} = 2.5 \, mol \).

Step3: Determine limiting reactant (using stoichiometry)

From reaction: \( 2 \, mol \, C_8H_{18} \) reacts with \( 25 \, mol \, O_2 \).
For \( C_8H_{18} \): Required \( O_2 = 0.3064 \, mol \times \frac{25}{2} = 3.83 \, mol \).
We have only \( 2.5 \, mol \, O_2 \), so \( O_2 \) is limiting.

Step4: Calculate moles of \( CO_2 \) from \( O_2 \)

From reaction: \( 25 \, mol \, O_2 \) produces \( 16 \, mol \, CO_2 \).
Moles of \( CO_2 = 2.5 \, mol \, O_2 \times \frac{16}{25} = 1.6 \, mol \).

Answer:

\( 1.6 \)