The amount of ${\mathrm{NaHCO}}_3$ required for obtaining 0.56 L of ${\mathrm{CO}}_2$ gas at STP is

$2{\mathrm{NaHCO}}_3\to {\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

$2 \mathrm{moles} \mathrm{of} {\mathrm{NaHCO}}_3\to 1 \mathrm{mole} \mathrm{of} {\mathrm{CO}}_2$

$2\times 84\mathrm{g} \mathrm{of} {\mathrm{NaHCO}}_3\to 22.4 \mathrm{lit} \mathrm{of} {\mathrm{CO}}_2$

$?\to 0.56 \mathrm{lit} \mathrm{of} {\mathrm{CO}}_2$

$22.4\times ?=2\times 84\times 0.56$

$=\frac{2\times {\cancel{84}}^{21}\times {\cancel{0.56}}^{0.1}}{{\cancel{22.4}}^4}$

$=4.2\mathrm{g} \mathrm{of} {\mathrm{NaHCO}}_3$