Answer:
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
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