8.7 of pure MnO2 is heated with an excess of HCl and the gas evolved is passed into a solution of KI. Calculate the weight of the iodine liberated (Mn = 55, CI = 35.5, I = 127)
Answers
Answered by
17
the amount of the iodine liberated (Mn = 55, Cl = 35.5, I =127)
0
MnO2 + 4HCl → Cl2 + MnCl2 + 2 H2O
One mole, i.e. 87g of MnO2 react to form one mole, i.e 71 g of Cl2.
So 8.7 g will form 7.1 g of Cl2.
Cl2 + 2KI → 2KCl + I2
One mole, i.e. 71 g of Cl2 react to form one mole, i.e 254 g of I2.
So 7.1 g will form 25.4 g of I2.
0
MnO2 + 4HCl → Cl2 + MnCl2 + 2 H2O
One mole, i.e. 87g of MnO2 react to form one mole, i.e 71 g of Cl2.
So 8.7 g will form 7.1 g of Cl2.
Cl2 + 2KI → 2KCl + I2
One mole, i.e. 71 g of Cl2 react to form one mole, i.e 254 g of I2.
So 7.1 g will form 25.4 g of I2.
Similar questions