Most commercial hydrochloric acid is prepared by heating NaCl with concentrated $H_{2}S{O}_4$. How much sulphuric acid containing $90.0\%$ $H_{2}S{O}_4$by mass is needed for the production of 200 kg of conc. HCl containing $42.0\%$ HCl by mass?

The correct option is A

126 kg

Since it is w-w problem, start by balancing the reaction.

2NaCl + $H_{2}S{O}_4$ $\to$ 2HCl + $N{a}_{2}S{O}_4$

1 mole of $H_{2}S{O}_4$ gives 2 moles of HCl
98g of $H_{2}S{O}_4$ gives (2 $\times$ 36.5)g of HCl
Since we require 200kg of conc. HCl having $42\%$ by mass of HCl HCl, $\frac{42}{100}$ $\times$ 200= 84kg of HCl
84kg of HCl = 112.77kg of $H_{2}S{O}_4$
But 112.77kg of $H_{2}S{O}_4$ is actually $90\%$ of the required weight of sulphuric acid
$\therefore$ 0.9x = 112.77

x = 126 kg