Question

consider the process below.
3fe₃o₄(s) + 8al(s) → 9fe(s) + 4al₂o₃(s)
a. what is the enthalpy of reaction for this process?
b. what amount of heat would be released when 1.000 × 10³ kg of iron is produced?
a. δhᵣ° = σ(nδhᵢ° products) - σ(nδhᵢ° reactants)
= (9 mol)(δhᵢ° fe(s)) + (4 mol)(δhᵢ° al₂o₃(s)) - (3 mol)(δhᵢ° fe₃o₄(s)) + (8 mol)(δhᵢ° al(s))
= (9 mol)(0 kj/mol)) + (4 mol)(-1675.7 kj/mol) - (3 mol)(-1118.4 kj/mol) + (8 mol)(0 kj/mol)
= (-6702.8 kj) + (3355.2 kj) = -3347.6 kj
b n_{fe(s)} = \frac{1000000 g}{55.85 g/mol} = 1.7905 × 10⁴ mol
\frac{-3347.6 kj}{9 mol fe(s)} = \frac{x}{1.7905 × 10⁴ mol fe(s)}
the quantity of heat given off would be 6.660 × 10⁶ kj

Explanation:

Step1: Calculate the amount of substance of Fe

Given \(m = 1.000\times10^{3}\text{ kg}=1.000\times10^{6}\text{ g}\), \(M = 55.85\text{ g/mol}\).
Using the formula \(n=\frac{m}{M}\), we have \(n_{Fe}=\frac{1.000\times 10^{6}\text{ g}}{55.85\text{ g/mol}}\approx1.7905\times 10^{4}\text{ mol}\)

Step2: Calculate the heat released

From the reaction \(3Fe_{3}O_{4}(s)+8Al(s)\to9Fe(s)+4Al_{2}O_{3}(s)\), \(\Delta H=- 3347.6\text{ kJ}\) (for the reaction producing \(9\text{ mol Fe}\)).
Let \(x\) be the heat released when \(n_{Fe} = 1.7905\times 10^{4}\text{ mol}\) is produced.
Set up the proportion \(\frac{-3347.6\text{ kJ}}{9\text{ mol Fe}}=\frac{x}{1.7905\times 10^{4}\text{ mol Fe}}\)
Cross - multiply: \(x=\frac{-3347.6\times1.7905\times 10^{4}}{9}\text{ kJ}\approx - 6.660\times 10^{6}\text{ kJ}\)

Answer:

The amount of heat released is \(6.660\times 10^{6}\text{ kJ}\)