Question

find the weight of 90% pure HCL required to react completely with 5 gram of 80% pure calcium​

Answer 1

100g of impure calcium carbonate contain :

=80g pure calcium carbonate

25g of impure calcium carbonate sample contain =

100

80

×25

=20g pure calcium carbonate

The desired equation is:

100g

1mol

CaCO

3

+2HCl→CaCl

2

+

atNTP

22.4litre

CO

2

+H

2

O

100gpure CaCO

3

liberate = 22.4 litre CO

2

20g pure CaCO

3

liberate =

100

22.4

×20

=4.48 litre CO

2

Hope this helps u

Answer 2

Answer:100g of impure calcium carbonate contain :

=80g pure calcium carbonate

25g of impure calcium carbonate sample contain =

100

80

​

×25

=20g pure calcium carbonate

The desired equation is:

100g

1mol

CaCO

3

​

​

​

+2HCl→CaCl

2

​

+

atNTP

22.4litre

CO

2

​

​

​

+H

2

​

O

100gpure CaCO

3

​

liberate = 22.4 litre CO

2

​

20g pure CaCO

3

​

liberate =

100

22.4

​

×20

=4.48 litre CO

2

​

Explanation: