Chemistry - Concentration Terms Question with Solution | TestHub

$\begin{array}{ccccccccc}{\text{Na}}_2{\text{CO}}_3& +& 2\text{HCl}& \to & 2\text{NaCl}& +& {\text{H}}_2\text{O}& +& {\text{CO}}_2\\ \text{1 mol}& & \text{2 mol}& & & & & & \end{array}$

${\text{Na}}_2{\text{CO}}_3=\text{1 g}=$ \frac{1}{106}\text{m}\text{o}\text{l}.\text{N}{\text{a}}_2\text{C}{\text{o}}_3\equiv \frac{2}{106}\text{m}\text{o}\text{l}.\text{H}\text{C}\text{l}$$

${\text{NaHCO}}_3+\text{HCl}\to \text{NaCl}+{\text{H}}_2\text{O}+{\text{CO}}_2$

${\text{NaHCO}}_3=\text{1 g}=\frac{1}{84}\text{mol}\equiv \frac{1}{84}mol\; HCl$

Total HCl required$=\left(\frac{1}{53}+\frac{1}{84}\right)mol\; HCl$

Let volume of 0.1 N HCl = V mL

then,$\text{Moles}= \frac{\text{0.1}\times \text{V}}{1000}=\frac{\text{V}}{10000}$

$\therefore \frac{\text{V}}{10000}=\left(\frac{1}{53}+\frac{1}{84}\right)$

$= 0.0308\; mol$

$\text{V}=\text{0.0308}\times 10000$

$= 308\; mL$