Question

30 g of NaOH is converted into NaCl and NaClO3 by the
action of Cly. The Cl2 is produced by reaction between
MnO2 and concentrated HCl. The amount of MnO2
required for the production of Cly is​

Answer 1

The mass of MnO2 produced is  30.375 g

Explanation:

Let the number of moles of MnO2 be x.

Now the reaction is

MnO2 + 4HCl -----> MnCl2 + Cl2 + 2H2O

The equation is

3Cl2 + 6 NaOH ----> 5 NaCl + NaClO3 + 3H2O

 x            2x                  

2x = 30 / 40

2x = 3 / 4

x = 3 / 8

Now no of moles of MnO2 = 3 / 8

Mass of MnO2 = 3/8 X 87

Mass of MnO2 =  30.375 g