Generally commercial hydrochloric acid is prepared by heating $N a C l$ with concentrated $H_{2} S O_{4}$ . How much $H_{2} S O_{4}$ solution containing 96 % $H_{2} S O_{4}$ by mass is required for the production 1000 kg of concentrated hydrochloric acid containing 43.8 % $H C l$ by weight)?

590.0 kg solution of

H_{2} S O_{4}

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310.0 kg solution of

H_{2} S O_{4}

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612.5 kg solution of

H_{2} S O_{4}

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588.0 kg solution of

H_{2} S O_{4}

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Solution

The correct option is C 612.5 kg solution of $H_{2} S O_{4}$
$2 N a C l + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H C l$
Mass of HCl = $1000 \times \frac{43.8}{100} = 438 k g$
$\frac{m o l e s o f H C l}{2} = \frac{m o l e s o f H_{2} S O_{4}}{1}$
$\frac{438 \times 10^{3}}{36.5 \times 2} = m o l e s o f H_{2} S O_{4}$
$M a s s o f H_{2} S O_{4} = \frac{98 \times 438 \times 10^{3}}{36.5 \times 2} = 588 \times 10^{3} g m$
Mass of 96 % $H_{2} S O_{4} = 588 \times \frac{100}{96} = 612.5 k g$

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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?