Question

20 g CaCo3 is treated with 20 g HCL . find limiting reagent and also . find the maximum amount of CO2 form​

Answer 1

Answer:

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.

Answer 2

Answer:

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