How many $mL$ of $0.1\, M\, HCl$ are required to react completely with $1\, g$ mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

(D) Let the mass of $Na_2CO_3$ be $x\, g$ and the mass of $NaHCO_3$ be $(1-x)\, g$.
Molar mass of $Na_2CO_3 = 106\, g\, mol^{-1}$ and $NaHCO_3 = 84\, g\, mol^{-1}$.
Since the mixture is equimolar,$\frac{x}{106} = \frac{1-x}{84}$.
Solving for $x$: $84x = 106 - 106x$ $\Rightarrow 190x = 106$ $\Rightarrow x = 0.5579\, g$.
Moles of $Na_2CO_3 = \frac{0.5579}{106} = 0.00526\, mol$.
Moles of $NaHCO_3 = \frac{1-0.5579}{84} = 0.00526\, mol$.
Chemical reactions:
$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$
$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$
Moles of $HCl$ required $= 2 \times (0.00526) + 1 \times (0.00526) = 0.01578\, mol$.
Volume of $0.1\, M\, HCl$ required $= \frac{0.01578\, mol}{0.1\, mol\, L^{-1}} = 0.1578\, L = 157.8\, mL$.