- 2.2g and 1.1g
- 1.1g and 2.5g
- 2.5g and 1.1g
- 5.0g and 1.1g
- 2.5g and 1.1g
Explanation:
Molecular mass of CaCO3: 40 + 12 + (16 × 3) = 100g
Molecular mass of CaO: 40 + 16 = 56g
Molecular mass of CO2: 12 + (16 × 2) = 44g
Reaction for the decomposition of calcium carbonate:
$\text{CaCO}_3\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{CaO}\ +\ \text{CO}_2\\ \ \ 100\text{g} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 56\text{g}\ \ \ \ \ \ \ \ 44\text{g}\\\ \ \ \ \text{X}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.4\text{g}\ \ \ \ \ \ \ \ \ \text{Y}$
56g of CaO is obtained when 100g of CaCO3 is burnt.
Therefore, to get 1.4g of CaO, the amount of CaCO3 to be burnt will be $\frac{100\ \times\ 1.4}{56}=2.5\text{g}$
For CO2,
100g of CaCO3 yields 44g of CO2
Therefore, 2.5g of CaCO3 will yield $\frac{44\ \times\ 2.5}{100}=1.1\text{g}$