Question

$$2 \text{ al}(s) + \text{fe}_2\text{o}_3(s) ightarrow 2 \text{ fe}(s) + \text{al}_2\text{o}_3(s) quad delta h^{circ} = -850 \text{ kj/mol}_{rxn}$$

the chemical equation shown above represents the thermite reaction. what is the approximate amount of heat released when 108 g of $\text{al}(s)$ reacts with excess $\text{fe}_2\text{o}_3(s)$ ?

a) 210 kj

b) 430 kj

c) 850 kj

d) 1700 kj

Explanation:

Step1: Calculate moles of Al

Molar mass of Al is 27 g/mol. Moles of Al = $\frac{108\ g}{27\ g/mol}$ = 4 mol.

Step2: Relate moles of Al to heat

From the reaction, 2 mol of Al releases 850 kJ. For 4 mol of Al, heat released = $\frac{4\ mol}{2\ mol} \times 850\ kJ$ = 1700 kJ? Wait, no, wait. Wait, the reaction is 2 Al gives -850 kJ. Wait, no, wait, 2 moles of Al react, ΔH is -850 kJ (released). So 2 moles Al → 850 kJ released. So 4 moles Al: (4/2)850 = 1700? But wait, no, wait the options: D is 1700? Wait, but let's check again. Wait, 108 g Al: 108/27 = 4 moles. The reaction is 2 Al(s) + Fe₂O₃(s) → 2 Fe(s) + Al₂O₃(s), ΔH = -850 kJ/mol rxn. So per 2 moles of Al, 850 kJ is released. So for 4 moles of Al, the heat released is (4/2)850 = 1700? But wait, the options have D as 1700 kJ. Wait, but maybe I made a mistake. Wait, no, 2 moles Al: 850 kJ. 4 moles Al: 2850 = 1700 kJ. So the answer should be D? Wait, but let's check the options again. The options are A 210, B 430, C 850, D 1700. So yes, D. Wait, but wait, maybe I messed up. Wait, 108 g Al is 4 moles. 2 moles Al give 850 kJ. So 4 moles give 2850 = 1700 kJ. So the answer is D.

Answer:

D. 1700 kJ