Question

1. how much energy is released during the formation of 98.7 g of fe according to the reaction below? fe2o3(s) + 2al(s) → al2o3(s) + 2fe(s) δrh° = -852 kj a) 753 kj b) 482 kj c) 241 kj d) 1.51 × 10^3 kj e) 4.20 × 10^3 kj

Explanation:

Step1: Calculate molar mass of Fe

The molar mass of Fe, $M_{Fe}=55.85\ g/mol$.

Step2: Calculate moles of Fe

The number of moles of Fe, $n=\frac{m}{M}$, where $m = 98.7\ g$. So $n=\frac{98.7\ g}{55.85\ g/mol}\approx1.77\ mol$.

Step3: Determine energy released

From the reaction $Fe_2O_3(s)+2Al(s)\to Al_2O_3(s) + 2Fe(s)$ with $\Delta_{r}H^{\circ}=- 852\ kJ$, for 2 moles of Fe formation, 852 kJ energy is released. For 1.77 moles of Fe, the energy released $E=\frac{1.77\ mol}{2\ mol}\times852\ kJ = 753\ kJ$.

Answer:

A. 753 kJ