Question
Chlorine gas is prepared by reaction of $${H_2}S{O_4}$$ with $$Mn{O_2}$$ and $$NaCl.$$ What volume of $$C{l_2}$$ 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 :
9.57$$\,L$$
Solution :
\[\underset{\begin{smallmatrix}
2\,moles \\
2\times 58.5=117\,g
\end{smallmatrix}}{\mathop{2NaCl}}\,+Mn{{O}_{2}}+3{{H}_{2}}S{{O}_{4}}\to 2NaHS{{O}_{4}}+MnS{{O}_{4}}+\underset{\begin{smallmatrix}
1\,mole \\
22.4\,L\left( STP \right)
\end{smallmatrix}}{\mathop{C{{l}_{2}}}}\,+2{{H}_{2}}O\]
$$117\,g\,\,{\text{of}}\,\,NaCl \equiv 22.4\,L\,\,{\text{of}}\,\,C{l_2}$$
$$50\,g\,\,{\text{of}}\,\,NaCl \equiv \frac{{22.4}}{{117}} \times 50$$ $$ = 9.57\,L\,\,{\text{of}}\,\,C{l_2}\,\,{\text{at}}\,\,STP$$