Question

Calculate the mass of residue obtained when $CaC{O}_3$ is strongly heated and $5.6\text{L}C{O}_2$ is produced at STP.

• A. $20\text{g}$
• B. $18\text{g}$
• C. $14\text{g}$
• D. $12\text{g}$

Answer 1

The correct option is C $14\text{g}$

$CaC{O}_3\left(s\right)⟶CaO\left(s\right)+C{O}_2\left(g\right)$

Moles of $C{O}_2$ at STP = $\frac{5.6}{22.4}$ = $0.25\text{mol}$

1 mol of $CaC{O}_3$ gives 1 mol of $CaO$ (residue) and 1 mol of $C{O}_2$.
So, when 0.25 mol of $C{O}_2$ is produced, 0.25 mol of residue will be obtained

weight of $CaO$ formed $=\text{Moles}\times \text{Molar mass}$
Hence, weight of $CaO=0.25\text{mol}\times 56\text{g/mol}=14\text{g}$