Question

A mass of $2.72$ g of an alloy containing $Pb,Al$ and $Cu$ was dissolved in $HN{O}_3$ containing $50\%$ by mass of $HN{O}_3$. When $H_{2}S{O}_4$ solution was added, $1.5$ g of a precipitate $\left(A\right)$ appeared. $H_{2}S$ gas was then passed into the remaining solution, a second precipitate was formed which when calcined in air produced $1.6$ g of a compound $B$. Determine the mass of $Al$ and composition of an alloy ($Al$ does not form $A{l}_2{S}_3$ due to the hydrolysis of sulphide).

• A. Mass of $Al=0.393g$
  $\%$ of $Pb=37.83$
  $\%$ of $Cu=46.98$
  $\%$ of $Al=15.91$
• B. Mass of $Al=0.593g$
  $\%$ of $Pb=30.83$
  $\%$ of $Cu=46.98$
  $\%$ of $Al=22.91$
• C. Mass of $Al=0.393g$
  $\%$ of $Pb=37.98$
  $\%$ of $Cu=41.83$
  $\%$ of $Al=20.91$
• D. None of these

Answer 1

The correct option is A Mass of $Al=0.393g$

$\%$ of $Pb=37.83$
$\%$ of $Cu=46.98$
$\%$ of $Al=15.91$
$Al,Pb,Cu\stackrel{H_{2}S{O}_4}{\to }\underset{\left(A\right)}{PbS{O}_4}$
$\underset{1mol}{Pb}\stackrel{HN{O}_3}{\to }Pb(N{O}_3{)}_2\stackrel{H_{2}S{O}_4}{\to }\underset{1mol}{PbS{O}_4}$
Mass of $Pb$ on passing $H_{2}S=\frac{1.5}{304}\times 208=1.026g$
$1$ mol $Cu=1$ mol of $CuS=1$ mol of $CuO$
Mass of $Cu=\frac{1.6}{79.5}\times 63.5=1.278g$
Mass of $Al=2.72-(1.026+1.278)=0.393g$
$\%$ of $Pb=37.83$
$\%$ of $Cu=46.98$
$\%$ of $Al=15.91$