A 5.72 g sample of impure cuprous oxide ( Cu 2 O ) is treated with dilute sulfuric acid, where Cu 2 O undergoes complete disproportionation to copper metal and CuSO 4 .The mixture is filtered to

Question

A 5.72 g sample of impure cuprous oxide $(Cu_{2} O)$ is treated with dilute sulfuric acid, where $Cu_{2} O$ undergoes complete disproportionation to copper metal and $CuSO_{4}$ .The mixture is filtered to remove the metallic precipitate, and the clear filtrate is treated with excess potassium iodide (Kl), producing a white precipitate and a brown solution.

This solution is again filtered, and excess H₂S is bubbled through the brown filtrate until the colour disappears, yielding 0.32 g of precipitate. Given atomic masses: H = 1, C = 6, N = 14, O = 16, S = 32, Cu = 63.5, K = 39, I = 127.

Calculate the percentage purity of the Cu₂O sample.

TestHub · Accepted Answer

1. Moles of sulfur: 0.32 g /32 = 0.0100 m o l . 2. From I 2 ​ + H 2 ​ S → 2 H I + S , moles of I 2 ​ = 0.0100 m o l . 3. From 2 C u 2 + + 4 I − → 2 C u I + I 2 ​ , moles of C u 2 + = 2 × 0.0100 = 0.0200 m o l . 4. Each mole of C u 2 ​ O gives 1 mole of C u 2 + ⇒moles C u 2 ​ O = 0.0200 m o l . 5. Mass of pure C u 2 ​ O = 0.0200 m o l × 143 g m o l − 1 = 2.86 g . 6. Purity = ( 2.86 g /5.72 g ) × 100 = 50% .

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