Cu ko phản ứng với H2SO4 loãng
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\Rightarrow m_{Cu}=4\left(g\right)\)
\(\Rightarrow\%Zn=\dfrac{6,5}{10,5}=62\%;\%Cu=100\%-62\%=28\%\)
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