How much amount of CuSO4·5H2O is required for liberation of 2.54 g of I2 when titrated with KI?<p></p>

2CuSO4·5H2O + 4KI → Cu2I2 + 2K2SO4 + I2 + 10H2O
Molecular weight of 2CuSO4·5H2O = [2(63.5 + 32 + 64) + 10(18)]g = 499 g
254 g of I2 is liberated by 499g CuSO4·5H2O.
2.54 g of I2 will be liberated by x g CuSO4·5H2O.
x = 499 × 2.54 / 254 = 4.99 g
Hence, the correct option is 4.99 g

Eduprobe

Question:

How much amount of CuSO4·5H2O is required for liberation of 2.54 g of I2 when titrated with KI?

Solution: