Question

1. formula 1 point

$2\ce{c6h14(l)} + 19\ce{o2(g)} \
ightarrow 12\ce{co2(g)} + 14\ce{h2o(g)}$ $\delta h_{\text{rxn}} = -4194\\ \text{kj}\\ \text{mol}^{-1}$
what is the energy change in kj released when 81.9 g of hexane, $\ce{c6h14}$, reacts with excess oxygen?
(note: this question has been updated. previous version wanted the quantity of energy; i have changed this to ask for the energy change.)
answer

Explanation:

Step1: Calculate molar mass of hexane

Molar mass of $\text{C}_6\text{H}_{14}$: $6\times12.01 + 14\times1.008 = 86.172\ \text{g/mol}$

Step2: Find moles of hexane

$$n = \frac{m}{M} = \frac{81.2\ \text{g}}{86.172\ \text{g/mol}} \approx 0.9423\ \text{mol}$$

Step3: Relate moles to energy change

Reaction releases 4194 kJ per 2 mol hexane. Energy per 1 mol: $\frac{4194}{2} = 2097\ \text{kJ/mol}$
Total energy released: $0.9423\ \text{mol} \times 2097\ \text{kJ/mol} \approx 1976\ \text{kJ}$

Answer:

The energy change (released) is approximately $\boldsymbol{1980\ \text{kJ}}$ (or $\boldsymbol{-1980\ \text{kJ}}$ if sign denotes direction, negative for exothermic)