Question

1.84g of a mixture contains caco3 and mgco3.On treating the mixture with dilute HCl .448ml of CO2 is obtained at STP. calculate the percentage composition of the mixture​

Answer 1

Answer:

Correct option is

A

50

Let CaCO

3

bexg.

Then, MgCO

3

is(18.4−x)g.

CaCO

3

→CaO+CO

2

1 mol 1 mol 1 mol

100g 56g 44g

xg

100

44x

g

MgCO

3

→MgO+CO

2

1 mol 1 mol 1 mol

84g 40g 44g

(18.4-x)g

84

44(18.4−x)

g

Total amount of CO

2

=

100

44x

+

84

44(18.4−x)

= 8.8 (given).

This gives x = 10g.

∴CaCO

3

=10g=0.1mol.

MgCO

3

=18.4−10=8.4g=0.1mol.

Total moles of CaCO

3

and MgCO

3

=0.2mol.

∴CaCO

3

=50mol%.

MgCO

3

=50mol%.