Question

Answer 1

a, Chỉ có Mg phản ứng với HCl, Cu không phản ứng

\(\rightarrow\left\{{}\begin{matrix}ddB:HCl,MgCl_2\\Y:Cu\end{matrix}\right.\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

          0,2<----------------------------0,2

\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(g\right)\\m_{Cu}=11,2-4,8=6,4\left(g\right)\end{matrix}\right.\\ \rightarrow n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

PTHH: \(Cu+2H_2SO_{4\left(đặc,nóng\right)}\rightarrow CuSO_4+SO_2\uparrow+2H_2O\)

           0,1------------------------------------------->0,1

\(b,\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{4,8}{11,2}.100\%=42,86\%\\\%m_{Cu}=100\%-42,86\%=57,14\%\end{matrix}\right.\)

\(d,n_{NaOH}=0,15.2=0,3\left(mol\right)\)

Xét \(T=\dfrac{0,3}{0,1}=3\) => Tạo muối Na₂SO₃ và NaOH dư

PTHH: \(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)

             0,2<-------0,1------>0,1

\(\rightarrow\left\{{}\begin{matrix}C_{M\left(NaOH.dư\right)}=\dfrac{0,3-0,2}{0,15}=\dfrac{2}{3}M\\C_{M\left(Na_2SO_3\right)}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\end{matrix}\right.\)