Chlorine gas is prepared by reaction of H₂SO₄ with MnO₂ and NaCl. What volume of Cl₂ will be produced at STP, if 50 g of NaCl is taken in the reaction?

$\underset{2 \mathrm{x} 58.5 = 117 \mathrm{g}}{\underset{2 \mathrm{moles}}{2\mathrm{NaCl}}}+{\mathrm{MnO}}_2+3{\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{NaHSO}}_4+{\mathrm{MnSO}}_4+\underset{22.4 \mathrm{L} \left(\mathrm{STP}\right)}{\underset{1 \mathrm{mole}}{{\mathrm{Cl}}_2}}+2{\mathrm{H}}_2\mathrm{O}$
117 g of NaCl $\equiv$ 22.4 L of Cl₂
50 g of NaCl $\equiv \frac{22.4}{117}\times 50=9.57 \mathrm{L} \mathrm{of} {\mathrm{Cl}}_2 \mathrm{at} \mathrm{STP}$