Chromium is one of the most abundant elements in Earth’s Crust and it is mined as chromite mineral, — Physical Chemistry — Kinetics Chemistry Question
Ferrochrome
Chromium is one of the most abundant elements in Earth’s Crust and it is mined as chromite mineral, FeCr2O4. South Africa, Kazakhstan, India, Russia, and Turkey are substantial producers. For the production of pure chromium, the iron has to be separated from the mineral in a two step roasting and leaching process.
4 FeCr2O4(s) + 8 Na2CO3(s) + 7 O2(g) → 8 Na2CrO4(s) + 2 Fe2O3(s) + 8 CO2(g)
2 Na2CrO4(s) + H2SO4(aq) → Na2Cr2O7(s) + Na2SO4(aq) + H2O(l)
Dichromate is converted to chromium(III) oxide by reduction with carbon and then reduced in an aluminothermic reaction to chromium.
Na2Cr2O7(s) + 2 C(s) → Cr2O3(s) + Na2CO3(s) + CO(g)
Cr2O3(s) + 2 Al(s) → Al2O3(s) + 2 Cr(s)
Calculate the mass of Cr that can be theoretically obtained from 2.1 tons of ore which contains 72.0 % of mineral FeCr2O4.
Model Answer
Mass of FeCr2O4 in the ore = 2.1 ⋅ 10^6 × 0.72 = 1.5 ⋅ 10^6 g
n(FeCr2O4) = 1.5 ⋅ 10^6 g / 224 g mol-1 = 6.7 ⋅ 10^3 mol
n(Cr) = 6.7 ⋅ 10^3 mol × 2 = 1.34 ⋅ 10^4 mol
m(Cr) = 1.34 ⋅ 10^4 mol × 52.0 g mol-1 = 7.0 ⋅ 10^5 g = 7.0 ⋅ 10^2 kg
Chromium, due to its strong corrosion resistance, is an important alloying material for steel. A sample of certain steel is to be analyzed for its Mn and Cr content. Mn and Cr in a 5.00 g steel sample are oxidized to MnO4 – and Cr2O7 2–, respectively, via a suitable treatment to yield 100.0 cm3 solution. A 50.0 cm3 portion of this solution is added to BaCl2 and by adjusting pH, chromium is completely precipitated as 5.82 g of BaCrO4. A second 50.0 cm3 portion of the solution requires exactly 43.5 cm3 of Fe2+ solution (c = 1.60 mol dm–3) for its titration in acidic solution. The unbalanced equations for the titration reactions are given below.
MnO4 – (aq) + Fe2+(aq) + H+(aq) → Mn2+(aq) + Fe3+(aq)
Cr2O7 2–(aq) + Fe2+(aq) + H+(aq) → Cr3+(aq) + Fe3+(aq)
Balance the equations for the titration reactions.
Model Answer
MnO4−(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)
Cr2O72−(aq) + 6 Fe2+(aq) + 14 H+(aq) → 2 Cr3+(aq) + 6 Fe3+(aq) + 7 H2O(l)
Calculate the % Mn and % Cr (by mass) in the steel sample.
Model Answer
n(BaCrO4) = 5.82 g / 253.3 g mol-1 = 2.30 ⋅ 10−2 mol
n(Cr2O7 2−) = 2.30 ⋅ 10−2 mol BaCrO4 × (1 mol Cr2O7 2- / 2 mol BaCrO4) = 1.15 ⋅ 10−2 mol
n(Cr) in 50.0 cm3 of the solution = 2.30 ⋅ 10–2 mol
n(Cr) in 100.0 cm3 of the solution = 4.60 ⋅ 10–2 mol
m(Cr) in 5.00 g of the steel sample = 4.60 ⋅ 10–2 mol × 52.0 g mol–1 = 2.39 g
n(Fe2+) used in the titration; 43.5 ⋅ 10–3 dm3 × 1.60 mol dm–3 = 6.96 ⋅ 10–2 mol
n(Fe2+) used for Cr2O7 2–: 1.15 ⋅ 10–2 mol × 6 = 6.90 ⋅ 10−2 mol
n(Fe2+) used for MnO4 – titration : (6.96 ⋅ 10−2 − 6.90 ⋅ 10−2 ) mol = 6 ⋅ 10−4 mol
n(Mn) in 50.0 cm3 of the solution = 6 ⋅ 10−4 × (1 mol MnO4– / 5 mol Fe2+) = 1.2 ⋅ 10−4 mol
n(Mn) in 100.0 cm3 of the solution = 2.4 ⋅ 10−4 mol
m(Mn) in 5.00 g of the steel sample = 2.4 ⋅ 10−4 mol × 54.9 g mol–1 = 0.013 g
% Mn = (0.013 g / 5.00 g) × 100 = 0.26
% Cr = (2.39 g / 5.00 g) × 100 = 48