Question

Chlorine gas is prepared by reaction of $H_{2}SO_{4}$ with $MnO_{2}$ and NaCl. What volume of $Cl_{2}$ will be produced at STP if 50 g NaCl is taken in the reaction?

• A. $1.915L$
• B. $22.4L$
• C. $11.2L$
• D. $9.57L$

Answer 1

Answer: (D)

$9.57L$

Answer 2

(4) : $\underset{\begin{aligned} & 2moles \\ & (2 \times 58.5 = 117g) \end{aligned}}{2NaCl} + MnO_{2} + 3H_{2}SO_{4} \rightarrow 2NaHSO_{4} + MnSO_{4} + \underset{\begin{aligned} & 1mle \\ & 22.4L(STP) \end{aligned}}{Cl_{2}} + 2H_{2}O$

117 g of NaCl $\equiv 22.4LofCl_{2}$

50 g of NaCl $\equiv \frac{22.4}{117} \times 50 = 9.57LofCl_{2}at$STP