Question
An amount of 0.15 moles of K2Cr2O7 is required to oxidize a mixture of XO and X2O3 (total mass = 25.56 g) to form XO4 − and Cr3+. If 0.21 moles of XO4 − is formed, then the correct information(s) is/are
Options
Solution
$5Cr_{2}O_{7}^{2 -} + 6XO + 34H^{+} \rightarrow 10Cr^{3 +} + 6XO_{4}^{-} + 17H_{2}O$
$x\text{ mole} \frac{6}{5}x\text{ mole} \frac{6}{5}x\text{ mole}$
$4Cr_{2}O_{7}^{2 -} + 3X_{2}O_{3} + 26H^{+} \rightarrow 8Cr^{3 +} + 6XO_{4}^{-} + 13H_{2}O $$${(0.15 - x)\text{ mole} \frac{3}{4}(0.15 - x)\text{ mole} \frac{6}{4}(0.15 - x)\text{ mole} }{\text{From question, }\frac{6}{5}x + \frac{6}{4}(0.15 - x) = 0.21 }{\Rightarrow x = 0.05 }{\text{Now},\frac{6}{5}x \times (A + 16) + \frac{3}{4}(0.15 - x) \times (2A + 48) = 25.56 \Rightarrow A = 100}$$
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