Question

The weight of carbon dioxide prepared by heating $200$ kg of Limestone which is $95\%$ pure is:

• A. $83.6$ kg
• B. $167.2$ kg
• C. $250.8$ kg
• D. $125.9$ kg

Answer 1

The correct option is A $83.6$ kg

${1CaCO}_3⟶1CaO+{1CO}_2{CaCO}_3⟶Limestone$

$1$mole of ${CaCO}_3$ gives $1$mole of CaO and $1$mol of ${CO}_2$

$100gof{CaCO}_3⟶1mole56gofCaO⟶1mole44gof{CO}_2⟶1mole100gof{CaCO}_3⟶44gof{CO}_{2}1gof{CaCO}_3⟶\frac{44}{100}gof{CO}_{2}200kgof{CaCO}_3⟶?\Rightarrow 95percentpure\frac{95}{100}\times 200\times 1000g=190\times {10}^{3}g190\times {10}^{3}gof{CaCO}_3⟶\frac{44}{100}\times 190\times {10}^{3}gof{CO}_2⟶83600gof{CO}_2\frac{83600}{1000}kgof{CO}_{2}83.6kgof{CO}_2$