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 $50\, g$ of $NaCl$ is taken in the reaction?

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

Answer 1

Answer: (D)

$\underset{\text{2$\,$moles(2$\times$ 58.5 = 117$\,$g)}}{2NaCl}$ $+MnO_{2}$$+3H_2SO_4 \to 2NaHSO_{4} +MnSO_{4} +$ $\underset{\text{1$\,$mole 22.4$\,$L(STP)}}{Cl_2}$ $+H_{2}O$
$117\,g$ of $NaCl \equiv 22.4\,L $ of $Cl_{2}$
$ 50 \,g$ of $NaCl \equiv \frac{22.4}{117} \times 50 = 9.57\, L$ of $Cl_{2}$ at $STP$