Question

A sample of magnesium metal containing some magnesium oxide as impurity was dissolved in 125 mL of 0.1 N $H_{2}S{O}_4$. The volume of hydrogen evolved at 273 and 1 atm was 120.1 mL. The resulting solution was found to be 0.02 N with respect to $H_{2}S{O}_4$. Calculate
(i) the weight of the sample dissolved and
(ii) the percentage by weight of magnesium in the sample. Neglect any change in the volume of the solution.

Answer 1

$Mg+MgO$
Dissolved in $H_2{SO}_4$ ($125ml$ and $0.1N$ $H_2{SO}_4$)
(1) $Mg+H_2{SO}_4\to Mg{SO}_4+H_2\uparrow$
$120.1ml$
AT $T={27.3}^{o}C,P=1$ atm
(2) $MgO+H_2{SO}_4\to Mg{SO}_4+H_{2}O$
$H_2{SO}_4$ remained $=0.02N$
$H_2{SO}_4$ reacted $=(0.1-0.02)=0.08N$
From equation (1)
mole of $H_2$ evolved $=$ mole of $Mg=$ mole of $H_2{SO}_4$ used in (1)
$\frac{120.1/1000\times 1}{0.821\times 300.3}=\frac{Wt.Mg}{24}$
Mole of $H_2{SO}_4=0.117g$
Mole of $H_2{SO}_4=4.871\times {10}^{-3}$ (Reacted in reaction (1))
moles of $H_2{SO}_4$ reacted in reaction (2)
$=$ Total moles reacted $-$ moles reacted reaction (1)
$=\frac{125}{1000}\times 0.08/2-4.871\times {10}^{-3}$
$=(5-4.871)\times {10}^{-3}=0.129\times {10}^{-3}$
Moles of $MgO$ in reaction (2) $=$ moles of $H_2{SO}_4$ reacted in (2)
$\frac{wt.MgO}{40}=0.129\times {10}^{-3}$
wt $MgO=0.00516g$
Total wt $(Mg-1000)0.1178g+0.00516g=0.123g$
% by wt of $Mg=\frac{0.1170}{0.123}\times 100=95.9$%