Question

How much limestone with 40% impurity can react with excess HCL to get 5.6 L of carbon dioxide gas​

Answer 1

ANSWER

CaCO3⇌ΔCaO+CO2

Let pure sample of CaCO3=x grams

Mass of CaO produced after decomposition =22.4 g

Molar mass of CaCO3=100 g/mol

and, Molar mass of CaO=56 g

If 100% is pure, then

100 g CaCO3⟶56 g of CaO

Also,y g of CaCO3⟶22.4 g of CaO

Dividing these two,

y100=22.456

y=40 grams

∴ Percentage of purity =TotalmassofimpureMassofpuresample×100=5040×100=80%

Answer 2

CaCO

3

⇌

Δ

CaO+CO

2

Let pure sample of CaCO

3

=x grams

Mass of CaO produced after decomposition =22.4 g

Molar mass of CaCO

3

=100 g/mol

and, Molar mass of CaO=56 g

If 100% is pure, then

100 g CaCO

3

⟶56 g of CaO

Also,y g of CaCO

3

⟶22.4 g of CaO

Dividing these two,

y

100

=

22.4

56

y=40 grams

∴ Percentage of purity =

Totalmassofimpure

Massofpuresample

×100=

50

40

×100=80%