Question

For the reaction : $Mn{O}_2+4HCl\stackrel{100\%}{\to }MnC{l}_2+C{l}_2+2H_{2}O$, $17.4gMn{O}_2$ is dissolved in $500mL$ of $HCl$ solution containing $7.3g$ $HCl$ per litre. (molar mass of $Mn=55g/mol$)

• A. $HCl$ is the limiting reagent
• B. $Mn{O}_2$ is the limiting reagent
• C. 0.05 moles of $MnC{l}_2$ will form
• D. 1120 mL $C{l}_2$ gas will be liberated at $0{}^{o}C$ and 1 atm

Answer 1

Answer: (A), (C), (D)

The correct options are
A $HCl$ is the limiting reagent
C 0.05 moles of $MnC{l}_2$ will form
D 1120 mL $C{l}_2$ gas will be liberated at $0{}^{o}C$ and 1 atm

$Mn{O}_2+4HCl\stackrel{100\%}{\to }MnC{l}_2+C{l}_2+2H_{2}O$

moles of $Mn{O}_2=\frac{17.4}{87}=0.2$
moles of $HCl=\frac{7.3}{36.5}=0.2$

$\frac{\text{moles of}Mn{O}_2}{\text{stoichiometric coefficeint}}=\frac{0.2}{1}=0.2$

$\frac{\text{moles of}HCl}{\text{stoichiometric coefficeint}}=\frac{0.2}{4}=0.05$

$HCl$ is a limiting reagent.

4 moles of $HCl$ form 1 mole of $C{l}_2$
$n_{C{l}_2}\text{formed from 0.2 mol of HCl}=0.05$
as molar volume at STP = $22.4L$
$V_{C{l}_2}=0.05\times 22.4=1120mL$
Similarly,$\text{moles of}MnC{l}_2=\frac{1}{4}\text{moles of HCl}$
$n_{MnC{l}_2}=0.05$