• # question_answer 44) Chlorine is prepared in the laboratory by treating manganese dioxide ($Mn{{O}_{2}}$) with aqueous hydrochloric acid according to the reaction: $4HCl(aq)+Mn{{O}_{2}}(s)\to MnC{{l}_{2}}(aq)$ $+2{{H}_{2}}O(l)+C{{l}_{2}}(g)$ How many gram of HCl react with 5g of manganese dioxide?

The given reaction is, $\underset{\underset{4\,\,\times \,36.5g}{\mathop{4\,mol}}\,}{\mathop{4HCl(aq)}}\,+\underset{\underset{\text{1 }\!\!\times\!\!\text{ 87g}}{\mathop{\text{1}\,\text{mol}}}\,}{\mathop{Mn{{O}_{2}}(s)}}\,\to \underset{1\,mol}{\mathop{MnC{{l}_{2}}(aq)}}\,+$ $\underset{2\,mol}{\mathop{2{{H}_{2}}O(l)}}\,+\underset{1\,mol}{\mathop{C{{l}_{2}}(g)}}\,$ $\because$$87gMn{{O}_{2}}$reacts with $4\times 36.5g\,HCl$ $\therefore$$5g\,Mn{{O}_{2}}$will react with $\frac{4\times 36.5}{87}\times 5g\,HCl$ = 8.39 g