Question

How many grams of copper will get replaced in \[{\rm{2}}\,\,{\rm{L}}\]of \[{\rm{CuS}}{{\rm{O}}_4}\] solution having molarity \[{\rm{1}}{\rm{.50}}\,\,{\rm{M}}\], if it is made to react with \[{\rm{54}}\,{\rm{ g}}\] of aluminium? (At. mass of \[{\rm{Cu}}\,\,{\rm{ = }}\,\,{\rm{63}}{\rm{.5}}\]and \[{\rm{Al}}\,\,{\rm{ = }}\,\,{\rm{27}}{\rm{.0}}\])

Answer 1

Hint: In mass-volume relationship problems, mass or volume of one of the reactants or products is calculated from the mass or volume of other substances.

Complete Step by Step Solution:
Molarity (M) may be defined as the number of moles of solute dissolved per litre of the solution.

Mathematically, \[{\rm{molarity}}\,{\rm{(M)}}\,{\rm{ = }}\dfrac{{{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{solute}}}}{{{\rm{volume}}\,\,{\rm{of}}\,\,{\rm{solution}}\,\,{\rm{in}}\,\,{\rm{litres}}}}\]
The units of molarity are moles per litre \[{\rm{(mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{)}}\]or moles per cubic decimetre \[{\rm{(mold}}{{\rm{m}}^{{\rm{ - 3}}}}{\rm{)}}\]. The symbol used to designate molar concentration is \[{\rm{M}}\].

The balanced molecular equation for the reaction of \[{\rm{CuS}}{{\rm{O}}_4}\]solution and aluminium may be represented as shown below:
\[{\rm{3CuS}}{{\rm{O}}_{\rm{4}}}\,{\rm{ + }}\,\,{\rm{2Al}}\,\, \to \,\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{(S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}_{\rm{3}}}\,\,{\rm{ + }}\,\,{\rm{3Cu}}\]

As per given data,
Volume of \[{\rm{CuS}}{{\rm{O}}_4}\, = \,{\rm{2}}\,\,{\rm{L}}\]
\[{\rm{CuS}}{{\rm{O}}_4}\, = \,{\rm{1}}{\rm{.50}}\,\,{\rm{M}}\,\,{\rm{or}}\,\,{\rm{1}}{\rm{.50}}\,\,\dfrac{{{\rm{mol}}}}{{\rm{L}}}\]

Find the moles of \[{\rm{CuS}}{{\rm{O}}_4}\]by using the molarity relationship as:
\[\begin{array}{l}{\rm{molarity}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_4}\,{\rm{ = }}\dfrac{{{\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_4}}}{{{\rm{volume}}\,\,{\rm{of}}\,\,\,{\rm{CuS}}{{\rm{O}}_4}\,\,{\rm{solution}}\,\,{\rm{in}}\,\,{\rm{litres}}}}\\ \Rightarrow {\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_4} = {\rm{molarity}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_4} \times \,{\rm{volume}}\,\,{\rm{of}}\,\,\,{\rm{CuS}}{{\rm{O}}_4}\,\,{\rm{solution}}\,\,{\rm{in}}\,\,{\rm{litres}}\\ \Rightarrow {\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_4} = {\rm{1}}{\rm{.50}}\,\,\dfrac{{{\rm{mol}}}}{{\rm{L}}} \times 2\,\,{\rm{L}}\\ \Rightarrow {\rm{number}}\,\,{\rm{of}}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_4} = {\rm{3}}\,\,{\rm{mol}}\end{array}\]
The relationship between moles (n), mass (m), and molar mass (M) is as shown below:
\[{\rm{moles}}\,{\rm{(n) = }}\dfrac{{{\rm{mass(m)}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}}}\]
moles of \[{\rm{CuS}}{{\rm{O}}_4}\,\] \[ = {\rm{3}}\,\,{\rm{mol}}\](calculated)
Molar mass of \[{\rm{CuS}}{{\rm{O}}_4}\,\]is known to be \[{\rm{159}}{\rm{.609}}\,\,\dfrac{{\rm{g}}}{{{\rm{mol}}}}\].
Find the mass of \[{\rm{CuS}}{{\rm{O}}_4}\,\]as shown below:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}}{\rm{ = }}\dfrac{{{\rm{mass(m)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}}}}\\ \Rightarrow {\rm{mass(m)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}} = {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}} \times {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}}\\ \Rightarrow {\rm{mass(m)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}} = \,\,{\rm{3}}\,\,{\rm{mol}} \times {\rm{159}}{\rm{.609}}\,\,\dfrac{{\rm{g}}}{{{\rm{mol}}}}\\ \Rightarrow {\rm{mass(m)}}\,\,{\rm{of}}\,\,{\rm{CuS}}{{\rm{O}}_{\rm{4}}} = \,478.8\,\,{\rm{g}}\end{array}\]

This means that \[478.8\,\,{\rm{g}}\]\[{\rm{CuS}}{{\rm{O}}_4}\,\]reacts with \[{\rm{54}}\,{\rm{ g}}\]of aluminium (\[{\rm{Al}}\,\])
Molar mass of copper (\[{\rm{Cu}}\]) is known to be \[{\rm{63}}{\rm{.5}}\,\,\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]

Also,
\[{\rm{159}}{\rm{.609}}\]\[{\rm{g}}\]\[{\rm{CuS}}{{\rm{O}}_4}\,\]contain \[{\rm{63}}{\rm{.5}}\,\,{\rm{g}}\,\,{\rm{Cu}}\]

So, \[{\rm{1}}\]\[{\rm{g}}\]\[{\rm{CuS}}{{\rm{O}}_4}\,\]will contain \[{\rm{ = }}\dfrac{{{\rm{63}}{\rm{.5}}\,\,}}{{{\rm{159}}{\rm{.609}}}}\,{\rm{g}}\,\,{\rm{Cu}}\]
\[\therefore \]\[478.8\,\,{\rm{g}}\]\[{\rm{CuS}}{{\rm{O}}_4}\,\]will contain \[{\rm{ = }}\dfrac{{{\rm{63}}{\rm{.5}}\,\,}}{{{\rm{159}}{\rm{.609}}}} \times 478.8\,\,{\rm{g}}\,\,{\rm{Cu}}\,\,{\rm{ = }}\,{\rm{190}}{\rm{.5}}\,\,{\rm{g}}\,\,{\rm{Cu}}\,\,\]
Hence, \[{\rm{190}}{\rm{.5}}\]grams of copper get replaced.
Therefore, the answer is \[{\rm{190}}{\rm{.5}}\,\,{\rm{g}}\,\,{\rm{Cu}}\]

Note: The only disadvantage of this concentration unit i.e., molarity is that its value changes with the change in temperature. It must be remembered that mass of one mole of atom is equal to the gram atomic mass of the element, mass of one mole of molecule is equal to gram molecular mass of the substance and mass of one mole of formula units in case of an ionic compound is equal to gram formula mass of the ionic compound.