Question

the mass of that can be prepared by heating 200 kg of limestone which is 95% pure is​

Answer 1

Explanation:

Amount of pure CaCO

3

=

100

95

×200=190kg =190000g

CaCO

3

(s)→CaO(s)+CO

2

(g)

1 mole CaCO

3

≡ 1 mole CaO

100 g CaCO

3

≡ 56 g CaO

∵ 100 g CaCO

3

give 56gCaO

∴ 190000 g CaCO

3

will give

100

56

×190000g=106400g=106.4kg