Question

The reduction of iron(iii) oxide (fe2o3) to pure iron during the first step of steelmaking, 2fe2o3(s)→ 4fe(s)+ 3o2(g) is driven by the high-temperature combustion of coke, a purified form of coal: c(s)+ o2(g)→ co2(g) suppose at the temperature of a blast furnace the gibbs free energies of formation δgf of co2 and fe2o3 are −429./kjmol and −835./kjmol, respectively. calculate the minimum mass of coke needed to produce 6400.kg of pure iron. round your answer to 2 significant digits.

Answer 1

Answer: 1031.33 kg of coke will produce 6400 kg of iron.

Explanation:

$2Fe_2O_3(s)\rightarrow 4Fe(s)+3O_2(g),\Delta G_{f}=-835kJ/mol$...(1)

$C(s)+O_2\rightarrow CO_2(g),\Delta G_{f}=-429kJ/mol$...(2)

(1)+3 × (2)

$2Fe_2O_3(s)+3C(s)\rightarrow 4Fe(s)+3CO_2(g),\Delta G_f=-1700kJ/mol$...(3)

$\Delta G_f=-835+3\times (-429)=-1700kJ/mol$

Mass of iron = 6400 kg = 6,400,000 g(1 kg = 1000 g)

Moles of iron =$\frac{\text{mass of the iron}}{\text{Molar mass of the iron}}=\frac{6,400,000 g}{55.85 g/mol}= 114,592.65 moles$

According to reaction (3) , 4 moles of iron are produced from 3 moles of coke(C(s)) ,then 114,592.65 moles of iron will be produced from = $\frac{3}{4}\times 114,592.65$ moles of coke.

Mass of coke:

$\text{Mass of coke}=\text{Number of moles of coke}\times \text{Molar mass of carbon}=\frac{3}{4}\times 114,592.65 mol\times 12 g/mol=1,031,333.93 g=1031.33 kg$

Hence, 1031.33 kg of coke will produce 6400 kg of iron.