\(1/\\ Fe+2HCl \to FeCl_2+H_2\\ Mg+2HCl \to MgCl_2+H_2\\ n_{HCl}=5.0,1=0,5(mol)\\ n_{Fe}=a(mol)\\ n_{Mg}=b(Mol)\\ m_{hh}=56a+24b=9,2(1)\\ n_{HCl}=2a+2b=0,5(2)\\ (1)(2)\\ a=0,1\\ b=0,15\\ \%m_{Fe}=\frac{0,1.56}{9,2}.100=60,86\%\\ \%m_{Mg}=100-60,86=39,13\%\)

\(V_{CH_4}=0,4.22,4=8,96(l)\)

\(\to B\\ Na_2SO_3+H_2SO_4 \to Na_2SO_4+SO_2+H_2O\)

\(n_{H_2SO_4}=n_{H_2}=a(mol)\\ BTKL:\\ m_{hh}+m_{H_2SO_4}=m_Y+m_{H_2}\\ 2,49+98.a= 8,25+2.a\\ \to a=0,06(mol)\\ a/ m_{H_2SO_4}=0,06.98=5,88(g)\\ b/ V_{H_2}=0,06.22,4=1,334(l)\)

\(n_{H_2SO_4}=n_{H_2}=a(mol)\\ BTKL:\\ m_{hh}+m_{H_2SO_4}=m_{muối}+m_{H_2}\\ 29+98.a=86,6+2.a\\ \to a=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\)

\(n_{CO_2}=\frac{5,6}{22,4}=0,25(mol)\\ n_{NaOH}=0,3.2=0,6(mol)\\ T=\frac{0,6}{0,25}=2,4\\ \Rightarrow Na_2SO_3\\ \Rightarrow m_{muối axit}=0(g)\\ \to Chọn A\)

\(Na+H_2O \to NaOH + \frac{1}{2}H_2\\ n_{Na}=\frac{4,6}{23}=0,2(mol)\\ n_{NaOH}=n_{Na}=0,2(mol)\\ CM_{NaOH}=\frac{0,2}{0,1}=2M\)

\(Na+HCl \to NaCl+\frac{1}{2}H_2\\ n_{Na}=2n_{H_2}=2.0,4=0,8(mol)\\ m_{Na}=0,8.23=18,4(g)\)

VHCl=0,1(l)=100ml mnh viết lộn =))

\(NaOH+HCl \to NaCl+H_2O\\ n_{NaOH}=0,2.1=0,2(mol)\\ n_{NaOH}=n_{HCl}=0,2(mol)\\ V_{HCl}=\frac{0,2}{2}=0,1M\)