Question

Which of the following statements is/are correct?
Commercial $HCl$ is prepared by heating $NaCl$ with $H_{2}S{O}_4$ as follows:

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

• A. $196.0$ g of pure $H_{2}S{O}_4$ is required for the production of $245.0$ g of conc. $HCl$ containing $40\%$ $HCl$ by weight.
• B. $245.0$ g of $80\%$ $H_{2}S{O}_4$ by weight is required for the production of $365.0$ g of conc. $HCl$ containing $40\%$ $HCl$ by weight.
• C. $2$ mol of pure $H_{2}S{O}_4$ is required for the production of $365$ g of $40\%$ $HCl$.
• D. $2.5$ mol of $80\%$ $H_S{O}_4$ is required for the production of $365.0$ g of $40\%$ $HCl$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B $245.0$ g of $80\%$ $H_{2}S{O}_4$ by weight is required for the production of $365.0$ g of conc. $HCl$ containing $40\%$ $HCl$ by weight.
C $2$ mol of pure $H_{2}S{O}_4$ is required for the production of $365$ g of $40\%$ $HCl$.
D $2.5$ mol of $80\%$ $H_S{O}_4$ is required for the production of $365.0$ g of $40\%$ $HCl$.

Using the following given equation, we can determine. $B,C$ and $D$ are correct.
$2NaCl+H_{2}S{O}_4\to N{a}_{2}S{O}_4+2HCl$

$B:$ Moles of $H_{2}S{O}_4=\frac{245\times 80}{100}\times \frac{1}{98}=2$ moles.

$365$ g of conc. $HCl$ containing $40$% $HCl$ by weight.

So, $HCl$ moles $=\frac{365\times 40}{100}\times \frac{1}{36.5}=4$ moles

As $1$ mol $H_{2}S{O}_4$ gives $2$ mol of $HCl$, so definitely $2$ mol $H_{2}S{O}_4$ will give $4$ mol of $HCl$.

$C:$ In the same way, $2$ mol of $H_{2}S{O}_4$ will give, $\frac{365\times 40}{100}\times \frac{1}{36.5}=4$ moles HCl.

$D:$ There are $2$ moles of pure $H_{2}S{O}_4$ in $2.5$ mol of $80$ % $H_{2}S{O}_4$.

So, it can produce only $\frac{365\times 40}{100}\times \frac{1}{36.5}=4$ moles $HCl$.