Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH:
\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)
\(Cu+HCl--\times-->\)
b. Theo PT(1): \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\Rightarrow\%_{m_{Mg}}=\dfrac{4,8}{11,2}.100\%=42,9\%\)
\(\%_{m_{Cu}}=100\%-42,9\%=57,1\%\)
c. Theo PT(1): \(n_{HCl}=2.n_{H_2}=2.0,2=0,4\left(mol\right)\)
PTHH: \(NaOH+HCl--->NaCl+H_2O\left(2\right)\)
Theo PT(2): \(n_{NaOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)