Question

80% purity of 200 g caco3 decomposes to produce the CO2,FIND the volume of co2 gas at Stp(approximately)

Answer 1

100 % pure CaCO3 => 100 g CaCO3 in 100g sample

80% pure CaCO3 => we have only 80g CaCO3 in 100g sample

Thus 200 g of 80% pure CaCO3 => 200 * (80/100) = 160g effectively

100 g CaCO3 => 1 mol

​160 g CaCO3 => 160/100 = 1.6 mole of CaCO3

Decomposition of 1 mole of CaCO3 gives 1 mole of CO2 gas ( 22.4 L at STP)

So 1.6 mole of CaCO3 decomposes to give 1.6 mole of CO2 gas ( 22.4 * 1.6 = 35.84 L at STP)