Question

10g of CaCO3 is treated with 250 ml of 2M HCI solution. Find the normality of HCl in resulting
solution.
please give answer with proper explanation.​

Answer 1

CaCO

3

+2HCl⟶H

2

CO

3

+CaCl

2

10gm250ml

In general 1 mole of CaCO

3

reacts with 2moles of HCl.

Number of moles of

M

CaCO

3

10

=

100

10

=0.1 mole.

Number of moles of HCl =Molarity×Volume=

1000

250

×1=0.25 mole.

0.1 mole of CaCO

3

will react with 0.2 moles of HCl(law of constant proportion).

So number of moles of HCl left =0.25−0.2=0.05 mole.

Number of equivalents of KOH =Number of equivalents of HCl

M

1

V

1

=M

2

V

2

2V

1

=0.05

V

1

=0.025l

_________

V=25ml