The volume of $C{O}_2$ obtained by the complete decomposition of $0.1mole$ of $NaHC{O}_3$ at STP is

$2NaHC{O}_3\underset{}{\overset{}{\to }}N{a}_{2}C{O}_3+C{O}_2+H_{2}O$

2 mole $NaHC{O}_3 \underset{}{\overset{}{\to }}22.4litC{O}_2$

$0.1moleNaHC{O}_3?$

$=\frac{22.4}{2}\times 0.1=1.12lit$