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 :
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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