PTHH: \(Mg+CuSO_4\rightarrow MgSO_4+Cu\)
Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\n_{CuSO_4}=\dfrac{250\cdot16\%}{160}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Mg còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=0,25\left(mol\right)=n_{MgSO_4}\\n_{Mg\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,25\cdot120}{7,2+250-0,05\cdot24-0,25\cdot64}=12,5\%\)
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