\begin{aligned}
&n_{A t}=\frac{1.56 g }{27 gmol ^{-1}}=0.05778 mol \\
&n_{C u}=x .
\end{aligned}harge on aluminium, \(C_{A l}=3\)harge on copper \(= C _{ Cu }=2\)rom Faraday's second law of electrolysis, \(n_{A l} C_{A l}=n_{ Cu }\) \(C_{ C v}\)
\(0.05778 \times 3=x \times 2\)
\begin{aligned}
x &=\left(\frac{0.05778 \times 3}{2}\right) mol \\
&=0.08667 mol
\end{aligned}
Mass of copper produced
\(=\) mole of copper \(x\) molar mass
\(=(0.08667 \times 63.5) g =5.5 g\)