p-Block elementsHard

Question

When 8.3 g copper sulphate reacts with excess of potassium iodide then the amount of iodine liberated is :

Options

A.42.3 g
B.24.3 g
C.4.23 g
D.2.43 g

Solution

2CuSO4. 5H2O + 4Kl → Cu2I2 + 2K2SO4 + I2 + 10H2O
                    498g                                           254g
498g of CuSO4 liberate I2 = 254g
8.3 g of CuSO4 liberate I2 = × 8.3
= 4.23 g on treatment with iodine produces :

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