Question

An impure sample of Calcium carbonate contains 80% pure calcium carbonate. 25g of impure sample reacted with excess of hydrochloric acid.Calculate the volume of carbon dioxide at NTPobtained from this sample.

Answer 1

The mass of pure Calcium carbonate is :

80/100 × 25 = 20g

Equation for the reaction is :

CaCO₃ (s) + 2HCl (aq) —> CaCl₂(s) + CO₂(g) + H₂O(l)

The mole ratio of Calcium carbonate and the CO₂ liberated is 1 : 1

Moles of Calcium carbonate = mass of calcium carbonate / molar mass

Molar mass of Calcium carbonate = 100

Moles = 20/100 = 0.2 moles

Moles of CO₂ = 0.2 moles

Molar gas volume at NTP is :

1 mol = 22.4 litres

0.2 moles =?

0.2 × 22.4 = 4.48 litres

CO₂ liberated is 4.48 litres

Answer 2

Answer:

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