A 5.72 g sample of impure cuprous oxide ( C u 2 O ) is treated with dilute sulfuric acid. During this process, C u 2 O undergoes complete disproportionation, forming copper metal and C u S O 4 .

Question

A 5.72 g sample of impure cuprous oxide ( $C u_{2} O$ ) is treated with dilute sulfuric acid. During this process, $C u_{2} O$ undergoes complete disproportionation, forming copper metal and $C u S O_{4}$ . The mixture is then filtered to remove the metallic precipitate. The clear filtrate is subsequently treated with excess potassium iodide ( $K I$ ), which results in a white precipitate and a brown solution.

This solution is filtered again. Excess $H_{2} S$ is then bubbled through the brown filtrate until its color disappears, yielding 0.32 g of precipitate. Given the 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 $C u_{2} O$ 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% .

Chemistry - Salt Analysis Question with Solution | TestHub

Doubts & Discussion