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,1.5g of a precipitate (A) appeared. \[{{H}_{2}}S\] gas was then passed into the remaining solution, a second precipitate was formed which when calculated in air produced 1.6g of a compound B. Determine the mass of Al and composition of alloy (Al does not form $A{{l}_{2}}{{S}_{3}}$ due to 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=37.83
        % of Cu=41.98
        % of Al=20.91

    (D) None of these

Answer 1

Hint: when Pb, Al and Cu are treated with sulphuric acid, precipitate is formed due to formation of Lead sulphate. On passing Hydrogen sulphide gas, copper sulphide is formed which on calcination produces cupric oxide. Composition can be calculated using the concept of stoichiometry.

Complete step by step answer:
When aluminum, lead and copper are treated with nitric acid, lead nitrate is formed which on reaction with sulphuric acid produces lead sulphate.
\[\begin{align}
  & Al,Pb,Cu\xrightarrow{{{H}_{2}}S{{O}_{4}}}PbS{{O}_{4}} \\
 & Pb\xrightarrow{HN{{O}_{3}}}Pb{{(N{{O}_{3}})}_{2}}\xrightarrow{{{H}_{2}}S{{O}_{4}}}PbS{{O}_{4}} \\
\end{align}\]
1 mole of lead produces 1 mole of lead nitrate which on reaction with sulphuric acid produces 1 mole of lead sulphate.
Mass of Pb on passing hydrogen sulphide gas=\[\dfrac{1.5}{304}\times 208\]=1.026g
Molecular mass of lead sulphate is 304 and Atomic mass of lead is 208.
-1 mole of copper produces 1 mole of copper sulphide on passing of hydrogen sulphide which on calcination will produce 1 mole of cupric oxide.

Mass of Cu=$\dfrac{1.6}{79.5}\times 63.5=$1.278g
Mass of Al=2.72-(1.026 + 1.278) =0.393 g
% of Pb=\[\dfrac{1.026}{2.72}\times 100=37.8%\]

Similarly, % of Cu =46.98% and % of Al =15.91%
So, the correct answer is “Option A”.

Note: Stoichiometry and mole ratios can be used to calculate compositions. Calcination indicates heating in presence of oxygen at high temperatures. Percent composition can be determined by dividing mass of that element to total mass. Molecular mass of Cupric oxide is 79.5