Question

6gm sample of CaCO3 reacts with 20gm solution of HCL having 20% by mass of HCL (density=1.10gm/ml). Calculate percentage purity of CaCO3

Answer 1

Answer:

8.8

Explanation:

CHEMISTRY

If 20 g of CaCO

3

is treated with 20 g of HCl, how many grams of CO

2

can be generated according to the following equation ?

CaCO

3

+2HCl→CaCl

2

+H

2

O+CO

2

(Molecular masses; CaCO

3

=100u,HCl=36.5uCO

2

=44u)

November 22, 2019avatar

Bhavinya Mishrikotkar

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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.

Answer 2

Answer:

The percentage purity of CaCO₃ is equal to 91.16%.

Explanation:

Weight of HCl in 20g of solution  $=20*\frac{20}{100} =4g$

Moles of HCl $=\frac{4}{36.5}=0.109moles$

         $CaCO_{3}+2HCl$  →  $CaCl_{2}+H_{2}CO_{3}$

moles of CaCO₃ reacted with HCl $=\frac{0.109}{2}=0.0547moles$

Molecular weight of CaCO₃ = 100g

Given Moles of CaCO₃  $=\frac{6}{100}=0.06moles$

% purity =( Reacted moles of CaCO₃/given moles of CaCO₃)×100

  Percentage purity $= (\frac{0.0547}{0.06})*100=91.16%$%