Question

If $20\mathrm{g}$ of $\mathrm{NaOH}$ and $36.5\mathrm{g}$ of $\mathrm{HCl}$ are added with $360\mathrm{g}$of water then weight of $\mathrm{HCl}$ present in resulting mixture is?

Answer 1

Step 1: Given data
Mass of $\mathrm{NaOH}=20\mathrm{g}$

Mass of $\mathrm{HCl}=36.5\mathrm{g}$

Mass of water$=360\mathrm{g}$

Step 2: Calculating the weight of $\mathrm{HCl}$

Moles of $\mathrm{NaOH}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}\Rightarrow \frac{20}{40}\Rightarrow 0.5\mathrm{mol}$

Moles of $\mathrm{HCl}=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{molecular}\mathrm{mass}}\Rightarrow \frac{36.5}{36.5}\Rightarrow 1\mathrm{mol}$

Volume of water$=0.36\mathrm{Litre}$

Writing a balanced equation for this reaction

$\underset{\mathrm{sodium}\mathrm{hydroxide}}{\mathrm{NaOH}\left(\mathrm{l}\right)}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{l}\right)}\to \underset{\mathrm{Sodium}\mathrm{chloride}}{\mathrm{NaCl}\left(\mathrm{l}\right)}+\underset{\mathrm{water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

$\mathrm{NaOH}$ is the limiting reagent in this reaction,

$0.5\mathrm{mol}$ of $\mathrm{NaOH}$ reacts with $0.5\mathrm{mol}$ of $\mathrm{HCl}$

A remaining mole of $\mathrm{HCl}$$=0.5\mathrm{mol}$

Weight of $\mathrm{HCl}=0.5\times 36.5\Rightarrow 18.25\mathrm{g}$

Mole of $\mathrm{NaCl}$ formed$=0.5\mathrm{mol}$

Weight of $\mathrm{NaCl}$$=0.5\times 58.5\Rightarrow 29.25\mathrm{g}$

Mole of water$=0.5$

Weight of water$=0.5\times 18\Rightarrow 9\mathrm{g}$

Total weight of the mixture$=18.25+29.25+9+360\Rightarrow 416.5\mathrm{g}$

Weight of $\mathrm{HCl}=\frac{18.25}{416.5}\times 100\Rightarrow 4.38\%$

Therefore, the weight of $\mathrm{HCl}=4.38\%$