Question

1.68 g of a mixture of dry sodium carbonate and sodium bicarbonate containing 75% of the latter was heated until there was no loss in weight.Calculate the volume of carbon dioxide at stp.
[Na=23 , C=12 ,O=16]

Answer 1

Sodium bicarbonate on decomposition gives sodium carbonate, carbon dioxide and water
$$\begin{gathered} 2{\mathrm{NaHCO}}_3\to {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O} \\ 2\mathrm{mol}1\mathrm{mol}1\mathrm{mol}=22.4\mathrm{L} \\ 2\times 84\mathrm{g}=168\mathrm{g} \end{gathered}$$
Sodium carbonate on heating gives sodium oxide and carbon dioxide as product.
$$\begin{gathered} {\mathrm{Na}}_2{\mathrm{CO}}_3\to {\mathrm{Na}}_2\mathrm{O}+{\mathrm{CO}}_2 \\ \mathrm{Since}{\mathrm{Na}}_2{\mathrm{CO}}_3\mathrm{is}\mathrm{also}\mathrm{produced}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{reaction},\mathrm{therefore},\mathrm{multiply}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{by}2 \\ \mathrm{so},\mathrm{the}\mathrm{reaction}\mathrm{will}\mathrm{be} \\ 2{\mathrm{Na}}_2{\mathrm{CO}}_3\to 2{\mathrm{Na}}_2\mathrm{O}+2{\mathrm{CO}}_2 \\ 2\mathrm{mol}2\mathrm{mol} \\ 2\times 106\mathrm{g}=212\mathrm{g}2\times 22.4\mathrm{L}=44.8\mathrm{L} \end{gathered}$$
Now, total amount of mixture= 1..68 g
Mass of NaHCO_​3 in the mixture= 75% of total mixture= 0.75 x 1.68 = 1.26 g
Mass of Na₂CO₃ in the mixture= 1.68g-1.26g = 0.42 g

Since, 168 g of NaHCO₃ gives 22.4 L of CO₂
Therefore, 1.26 g of NaHCO₃ gives
$\frac{22.4}{168}\times 1.26=0.168\mathrm{L}$
And,
212 g of Na₂CO₃ gives 44.8 L of CO₂
So, 0.42 g of Na₂CO₃ gives
$\frac{44.8}{212}\times 0.42=0.088\mathrm{L}$
Total volume of CO₂ produced = 0.168 L + 0.088L= 0.256 L