Question

The volume of $C{O}_2$ (in $c{m}^3$) liberated at S.T.P. when $1.06$ g of anhydrous sodium carbonate is treated with an excess of dilute HCl is:

• A. $112$
• B. $224$
• C. $56$
• D. $2240$

Answer 1

The correct option is B $224$

1.06 g of sodium carbonate correspond to $\frac{1.06}{106}=0.01$ moles.

$N{a}_{2}C{O}_3+2HCl\to 2NaCl+H_{2}O+C{O}_2$

1 mole of sodium carbonate liberates 1 mole of $C{O}_2$.

0.01 moles of sodium carbonate will liberate 0.01 moles of $C{O}_2$.

1 mole of $C{O}_2$ at STP occupies a volume of $22400c{m}^3$.

$0.01$ mole of $C{O}_2$ at STP will occupy a volume of $224c{m}^3$.