Question

Chlorine is prepared in the laboratory by treating manganese dioxide ($Mn{O}_2$) with aqueous hydrochloric acid according to the reaction
4 HCl (aq) + $Mn{O}_2\left(s\right)\to 2H_{2}O\left(l\right)+MnC{l}_2\left(aq\right)+C{l}_2\left(g\right)$
How many grams of $HCl$ react with 5.0 g of manganese dioxide?

Answer 1

According to given reaction,

$4$ moles of $HCl$ reacts with $1$ mole of $Mn{O}_2$

Given moles of $Mn{O}_2=\frac{5}{86.94}=0.0575$

$\therefore$ Moles of $HCl=4\times 0.0575=0.23$

$\therefore$ Mass of $HCl$ reacting= $0.23\times 36.5=8.395g$ .