Question

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$

If $1$ g of $HCl$ and $1$ g of $Mn{O}_2$ is heated together, the maximum weight of $C{l}_2$ gas evolved will be:

• A. $2$ g
• B. $0.975$ g
• C. $0.486$ g
• D. $0.972$ g

Answer 1

The correct option is C $0.486$ g

The molar mass of $HCl$ and $Mn{O}_2$ are $36.5$ g/mol and $87$ g/mol respectively.
$1$ g of $HCl$ corresponds to $\frac{1}{36.5}=0.027$ moles, while $1$ g of $Mn{O}_2$ corresponds to $\frac{1}{87}=0.0115$ moles.

$1$ mole of $Mn{O}_2$ reacts with $4$ moles of $HCl$.
$0.0115$ moles of $Mn{O}_2$ will react with $0.046$ moles of $HCl$ but only $0.027$ moles of $HCl$ are present.
Hence, $HCl$ is the limiting reagent.
$4$ moles of $HCl$ will produce $1$ mole of chlorine.

Hence, $0.027$ moles of $HCl$ will produce $0.25\times 0.027=0.00675$ moles of chlorine which corresponds to $0.00675\times 71=0.486$ g.