Question

the (balanced) exothermic reaction shown below is used by railroad crews to weld rails together:
2 al(s) + fe₂o₃(s) → 2 fe(s) + al₂o₃(s) δh = -850. kj
what mass of aluminum oxide mw = 101.97 g mol⁻¹ is formed if 236 kj of heat are released?
○ 52.8 g
○ 28.3 g
○ 160. g
○ 0.278 g

Explanation:

Step1: Find moles of reaction from heat

The reaction has $\Delta H = -850. \text{kJ}$ (per 1 mol of reaction, since coefficients are for 1 reaction). Heat released is 236 kJ (exothermic, so $\Delta H$ is negative, but we use magnitude for moles).
Moles of reaction, $n_{\text{rxn}} = \frac{\text{Heat released}}{|\Delta H|} = \frac{236 \text{ kJ}}{850. \text{ kJ/mol-rxn}}$
$n_{\text{rxn}} = \frac{236}{850} \approx 0.2776 \text{ mol-rxn}$

Step2: Moles of $\text{Al}_2\text{O}_3$ from reaction stoichiometry

From the balanced equation: $1 \text{ mol Al}_2\text{O}_3$ is formed per $1 \text{ mol-rxn}$.
So, $n_{\text{Al}_2\text{O}_3} = n_{\text{rxn}} = 0.2776 \text{ mol}$

Step3: Mass of $\text{Al}_2\text{O}_3$ from moles and molar mass

Molar mass (MW) of $\text{Al}_2\text{O}_3 = 101.97 \text{ g/mol}$.
Mass, $m = n \times \text{MW} = 0.2776 \text{ mol} \times 101.97 \text{ g/mol}$
$m \approx 0.2776 \times 101.97 \approx 28.3 \text{ g}$

Answer:

28.3 g