Question

Q.20g of calcium carbonate on treating with
20g HCl forms carbon dioxide ?
I. find out the limiting reagent ?
2. Name the excess reagent & determine the
amount of excess reagent ?
3. calculate Maximum amount of carbon
diexide formed ?​

Answer 1

Answer:

ANSWER

100g

1mol

CaCO

3

(s)

+

73g

2mol

2HCl(aq.)

→CaCl

2

(aq.)+H

2

O(l)+

44g

1mol

CO

2

(g)

Let CaCO

3

(s) be completely consumed in the reaction.

∵100gCaCO

3

give 44gCO

2

∴20gCaCO

3

will give

100

44

×20gCO

2

=8.8gCO

2

Let HCl be completely consumed.

∵ 73 g HCl give 44 g CO

2

∴ 20 gHCl will give

73

44

×20gCO

2

=12.054gCO

2

Since, CaCO

3

gives least amount of product CO

2

, hence CaCO

3

is limiting reactant. Amount of CO

2

formed will 8.8 g