Ta có: \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Theo PT: nMg = nH2 = 0,3 (mol)
⇒ mMg = 0,3.24 = 7,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{15}.100\%=48\%\\\%m_{Cu}=100-48=52\%\end{matrix}\right.\)