Question

$1.2g$ of $Mg$ is burnt in a closed vessel which contains $0.5g$ of $O_2$. Millilitres of $0.25M{H}_2{S}_4$ required to dissolve the residue in the vessel is

• A. $40mL$
• B. $175mL$
• C. $200mL$
• D. $250mL$

Answer 1

Answer: (A)

$40mL$

Given,

$1.2g$ $Mg$ burnt with $0.5g$ of $O_2$

The reaction involved is:-

$2Mg+O_2⟶2MgO$

It is clear that $2$ moles of $Mg$ react with $1$ mole of $O_2$ to produce $2$ moles $MgO$

Now, Moles of given $Mg=\frac{1.2}{24}=0.05$ moles

Also, Moles of gain $O_2=\frac{0.5}{32}=0.015\approx 0.02$ moles

$\therefore O_2$ here is limiting reagent and $0.05$ moles of $Mg$ require $0.025$ moles of $O_2$ but $O_2$ is only $0.02$ moles given.

$\Rightarrow 0.02$ moles of $O_2$ will burn $2\times 0.02=0.04$ moles of $Mg$ to give $0.04$ moles of $MgO$

$\Rightarrow$ Moles of $Mg$ remained in vessel=$0.05-0.04=0.01$ moles

Now, the next reaction involved is

$Mg+H_{2}S{O}_4⟶MgS{O}_4+H_2$

Now, it is clear that $1$ mole $Mg$ consume $1$ mole $H_{2}S{O}_4$

$\Rightarrow 0.01$ moles of $Mg$ will consume $0.01$ mole $H_{2}S{O}_4$

But we are given with $0.25M$ $H_{2}S{O}_4$

$\therefore$ Volume of $H_{2}S{O}_4$ required, which has molarity= $0.25M$ & No. of moles=$0.01$

Volume=$\frac{No.ofmoles}{Molarity}$

$=\frac{0.01}{0.25}$

$=0.04L$

$\therefore$ No. of millilitres of $0.25M$ $H_{2}S{O}_4$ required=$40ml$ [$\because 0.04=40\times {10}^{-3}$]