(a)
$2NaHCO_{3}\overset{\mathrm{\Delta}}{\rightarrow}Na_{2}CO_{3} + H_{2}O + CO_{2}$
2 mol 1 mol
$\because$ 2 mole $NaHCO_{3}$ on decomposition gives= 1 moles $Na_{2}CO_{3}$
$\therefore$ 0.2 mole $NaHCO_{3}$ on decomposition will give
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{1}{2} \times 0.2$
=0.1 mol $Na_{2}CO_{3}$