Question

How much amount of CaCO3 in gram having percentage purity 50 percent produces 0.56 litres of CO2 at STP on heating?
Please give full solution.

Answer 1

Answer: 5.0 grams

Explanation: $CaCO_3\rightarrow CaO+CO_2$

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and weighs equal to the molecular mass.

From the balanced chemical equation:

100 grams of $CaCO_3$ decomposes to give 22.4 L  of $CO_2$  at STP.

22.4 L of $CO_2$ are produced from 100 grams of $CaCO_3$

0.56 L of $CO_2$ will be produced from=$\frac{100}{22.4}\times 0.56=2.5grams$ of $CaCO_3$

But as percentage purity is 50% , the amount of $CaCO_3$ required is $2.5\times \frac{100}{50}=5.0 grams$

Answer 2

Answer:

Answer: 5.0 grams

Explanation: CaCO_3\rightarrow CaO+CO_2CaCO3→CaO+CO2