Question

How many grams of copper will be replaced in $2\, L$ of a $1.50- M\, CuSO_4$ solution if the latter is made to react with $27.0\, g$ of aluminium ? $( Cu =63.5, Al =27.0 )$
$\left(3 CuSO _{4}+2 Al \rightarrow Al _{2}\left( SO _{4}\right)_{3}+3 Cu \right)$

Answer 1

$\frac{W_{1}}{W_{2}}=\frac{E_{1}}{E_{2}}$
$\frac{W_{1}}{27}=\frac{63.5}{2} \times \frac{3}{27}$
$W_{1}=63.5 \times \frac{3}{2} \times \frac{27}{27}=95.25\, g$