How many grams of copper will be replaced in $2\mathrm{L}$ of a $1.50 \mathrm{M} {\mathrm{CuSO}}_4$ solution if the latter is made to react with $27.0$of aluminium $(Cu=63.5,Al=27.0)$

$2\mathrm{L}\text{ of }1.5\mathrm{M}{\mathrm{CuSO}}_4=3\mathrm{mol}\text{ of }{\mathrm{CuSO}}_4$

$\begin{array}{l}2\mathrm{Al}+3{\mathrm{CuSO}}_4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\to {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+3\mathrm{Cu}\\ 2\mathrm{mol}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\mathrm{mol} 3 \mathrm{mol}\end{array}$

$$\begin{gathered} 2\mathrm{mol}\text{ of }\mathrm{Al}=3\mathrm{mol}\text{ of }\mathrm{Cu} \\ (27g=1\mathrm{mol}\text{ of }Al) \\ =\frac{3}{2}\times 63.5\mathrm{g}\text{ of }\stackrel{ }{C}u=95.25\mathrm{g} \end{gathered}$$