Step 1 : Calculation of the number of moles of the components in the mixture.Let $\mathrm{Na_2CO_3}$ present in the mixture $= x \ \mathrm{g}$ $\therefore \; \mathrm{NaHCO_3}$ present in the mixture $= (1 - x) \ \mathrm{g}$Molar mass of $\mathrm{Na_2CO_3}$ $= 2 \times 23 + 12 + 3 \times 16$ $= 106 \ \mathrm{g\,mol}^{-1}$Molar mass of $\mathrm{NaHCO_3}$ $= 23 + 1 + 12 + 3 \times 16$ $= 84 \ \mathrm{g} \ \mathrm{mol}^{-1}$ $\therefore \;$ No. of moles of $\ \mathrm{Na_2CO_3}$ in $x \ \mathrm{g} = \frac{x}{106}$No. of moles of $\ \mathrm{NaHCO_3} \ \text{in}\ (1 - x) \ \mathrm{g} = \frac{1 - x}{84}$ As mixture contains equimolar amounts of the two, $\frac{x}{106} = \frac{1 - x}{84}$ or, $106 - 106x = 84x$ or $x = \frac{106}{190} \ \mathrm{g} = 0.558 \ \mathrm{g}$Thus, moles of $\ \mathrm{Na_2CO_3} = \frac{0.558}{106} = 0.00526$ $\therefore \;$ Moles of $\mathrm{NaHCO_3} = 0.00526$Step 2 : To calculate the moles of HCl required. $\mathrm{Na_2CO_3} + 2 \mathrm{HCl} \rightarrow 2 \mathrm{NaCl} + \mathrm{H_2O} + \mathrm{CO_2}$ $\ \mathrm{NaHCO_3} + \ \mathrm{HCl} \rightarrow \mathrm{NaCl} + \ \mathrm{H_2O} + \ \mathrm{CO_2}$1 mole of $\ \mathrm{Na_2CO_3}$ requires $\ \mathrm{HCl} = 2$ moles $\therefore \; 0.00526$ mole of $\ \mathrm{Na_2CO_3}$ requires $\ \mathrm{HCl} = 0.00526 \times 2 = 0.01052 \ \mathrm{mole}$1 mole of $\mathrm{NaHCO_3}$ requires $\ \mathrm{HCl} = 1 \ \mathrm{mole}$ $\therefore \; 0.00526$ mole of $\ \mathrm{NaHCO_3}$ requires $\ \mathrm{HCl} = 0.00526 \ \mathrm{mole}$$\therefore \;$ Total $\ \mathrm{HCl}$ required $= 0.01052 + 0.00526 = 0.01578 \ \mathrm{mole}$Step 3 : To calculate volume of 0.1 M HClVolume of acid $= \frac{ \text{Number of moles} }{ \text{Molarity} }$ $= \frac{0.01578}{0.1} = 157.8 \ \mathrm{mL}$