Question

If 25 g each of CaCO3 and H2SO4 reacts with each other, the amount of CaSO4 formed
A) 30 g C) 34 g B) 32 g D) 36 g​

Answer 1

Answer:

hope it will help you dont worry its 10000000 percent correct

Explanation:

Solution:-

Molecular weight of CaCO

3

=100g

Molecular weight of CO

2

=44g

Decomposition of CaCO

3

-

CaCO

3

⟶CaO+CO

2

Now, from the above reaction-

Weight of pure CaCO

3

required to produce 44g of CO

2

=100g

Weight of pure CaCO

3

required to produce 8g of CO

2

=

44

100

×8≈18.18g

Given weight of CaCO

3

=25g

Now,

Amount of pure CaCO

3

in 25g of given MgCO

3

=18.18g

Thus,

Amount of pure CaCO

3

in 100g of given MgCO

3

=

25

18.18

×100=72.72g

Therefore,

The percentage purity of given CaCO

3

=72.72%