Question

4. How much carbon dioxide i liberated when 50gm of calcium carbonate reacts with excess of hydrochloric acid.ca=40 U, C = 120,0 = 16u, H = 0,1u = 35.5u)
caco3(s )+2hcl( aq)➡️cacl2( 2a)+co2(g )​

Answer 1

CaCO

3

+2HCl⟶CaCl

2

+H

2

CO

3

HCl is limiting reagent so the moles of H

2

CO

3

=10

−1

mol

Now-

H

2

CO

3

⟶H

2

O+CO

2

n mole of H

2

CO

3

will give same amount of mole for CO

2

The no. of mole of CO

2

=10

−1

mole

The volume of CO

2

-

mole=

22.4

Volume

10

−1

=

22.4

volume

Volume=2.24L

i hope it helps you...