Question

Calculate the weight of 80% pure CaCO3 to react completely 25 g of H2SO4. Also calculate

the volume of CO2 gas formed at NTP.​

Answer 1

The answer are- 31.25 grams of CaCO₃ and 22.4 liters of CO₂ gas at NTP

GIVEN

80% pure CaCO₃ to react completely 25 g of H₂SO₄

TO FIND

Weight of 80% pure CaCO₃ and volume of CO2 gas formed at NTP.

SOLUTION

The above problem can be simply solved as follows;

Reaction between CaCO₃ and H₂SO₄

CaCO₃ + H₂SO₄ ----------> CaSO₄ + H₂O + CO₂

1 mole of H₂SO₄ reacts with 1 mole of calcium carbonate.

Mass of 1 mole of H₂SO₄ = 98 grams

Given mass = 25 grams

Moles of H₂SO₄ = 25/98 = 0.25 moles

Moles of Calcium carbonate = 0.25 moles

Mass = 0.25 × 100 = 25 grams

Let the mass of 80% pure CaCO₃ = x

80% of x = 25

0.8x = 25

x = 31.25grams

Hence, The answer is 31.25 grams of CaCo₃

moles of CO₂ formed = 1

Volume of 1 mole of CO₂ at NTP = 22.4 liters

Hence, The answer are- 31.25 grams of CaCO₃ and 22.4 liters of CO₂ gas at NTP

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