Question

When 0.36g of Iron pyrite (FeS₂) was heated strongly In alr following reaction takes place FeS₂(s) + O₂(g) → SO₂(g) + Fe₂O₃(s) The SO₂(g) produced was titrated with acidicfied K₂Cr₂O₇ solution. Calculate the volume In mL of 1M K₂Cr₂O₇ solution used In redox titration. (MW of Fe = 56 and S = 32)

• A. 2 mL
• B. 4 mL
• C. 6 mL
• D. 8 mL

Answer 1

Answer: (A)

2 mL

Answer 2

1. Balance the reaction: $4FeS_2(s) + 11O_2(g) \rightarrow 8SO_2(g) + 2Fe_2O_3(s)$.
2. Calculate moles of $FeS_2$: $MW(FeS_2) = 56 + 2(32) = 120$ g/mol. Moles = $0.36g / 120 g/mol = 0.003$ mol.
3. Moles of $SO_2$ produced: From the balanced equation, 4 moles $FeS_2$ produce 8 moles $SO_2$. So, 0.003 mol $FeS_2$ produces $2 \times 0.003 = 0.006$ mol $SO_2$.
4. Redox reaction: $3SO_2 + Cr_2O_7^{2-} \rightarrow 3SO_4^{2-} + 2Cr^{3+}$. The mole ratio of $SO_2$ to $K_2Cr_2O_7$ is 3:1.
5. Moles of $K_2Cr_2O_7$ required: Moles $K_2Cr_2O_7 = 0.006$ mol $SO_2 \times (1 \text{ mol } K_2Cr_2O_7 / 3 \text{ mol } SO_2) = 0.002$ mol.
6. Volume of $K_2Cr_2O_7$: Volume = Moles / Molarity = $0.002$ mol / 1 M = $0.002$ L = 2 mL.