How much amount of CuSO ₄ 5 H ₂ O is required for liberation of 2.54 ~g of I ₂ when titrated with KI ?

Step-by-step solution

aligned & 2 CuSO ₄ 5 H ₂ O +4 KI & Cu ₂ I ₂+2 ~K ₂ SO ₄+ I ₂+10 H ₂ O & 254 ~g & aligned Molecular weight of 2 CuSO ₄ 5 H ₂ O [2(63.5+32+64)+10(18)] g =499 ~g 254 ~g of I ₂ is liberated by 499 ~g CuSO 5 H ₂ O 2.54 ~g of I ₂ will be liberated by x g CuSO₄ 5 H ₂ O x= 499 254 2.54=4.99 ~g