\(Fe(NO_3)_3\\ \%_{Fe}=\frac{56}{242}.100\%=23,14\%\\ \%_N=\frac{14.3}{242}.100\%=17,36\%\\ \%_O=595,5\%\\ (NH_4)_2SO_4\\ \%N=\frac{14.2}{132}=21,21\% \\ \%H=\frac{(1.4).2.}{132}=6,06\% \\ \%S=\frac{32}{132}=24,24\% \\ \%O=\frac{16.4}{132}=48,49\%\\ \)

\(V_{ddHCl}=\frac{0,5}{2}=0,25(l)=250(ml)\)

\(n_{NaOH}=\frac{60}{40}=1,5(mol)\\ V_{ddNaOH}=\frac{1,5}{5}=0,3(l)=300(ml)\)

\(Na+HCl \to NaCl+\frac{1}{2}H_2O\\ n_{Na}=\frac{6,9}{23}=0,3(mol)\\ n_{HCl}==0,4.1=0,4(mol)\\ 0,3<0,4\\ \Leftrightarrow Na < HCl\\ CM_{NaCl}=\frac{0,3}{0,4}=0,75M\)

\(\text{Chọn C: Ag}\\ 2Ag +2H_2SO_4 \buildrel{{t^o}}\over\longrightarrow Ag_2SO_4+SO_2+2H_2O\)

\(\text{Chọn A}\\ H_2: 2 (đvc)\\ O_2: 32(đvc)\\ HCl: 36,5 (đvc)\\ N_2: 28 (đvc) \)

\(\text{Chọn B:}\\ \text{A sai do HCl là axit}\\ \text{C sai do Cu nằm sau H ko phản ứng}\\ \text{D sai do Ag nằm sau H ko phản ứng}\\ \)

\(Zn+2HNO_3 \to Zn(NO_3)_2+H_2O\\ n_{Zn}=\frac{13}{65}=0,2(mol)\\ n_{HNO_3}=2.n_{Zn}=2.0,2=0,4 (mol)\\ V_{HNO_3}=\frac{0,4}{0,1}=4(l)=4000(ml)\)

\(Mg+2HCl \to MgCl_2+H_2\\ Fe+2HCl \to FeCl_2+H_2\\ n_{Mg}=a(mol)\\\ n_{Fe}=b(mol) m_{hh}=24a+56b=5,2(1)\\ n_{HCl}=2a+2b=0,3(2)\\ (1)(2) a=0,1;b=0,05\\ \%m_{Mg}=\frac{0,1.24}{5,2}.100\%=46,15\%\\ \%m_{Fe}=53,85\%\)

\(CuO+H_2 \to Cu+H_2O\\ Fe_2O_3+3H_2 \to 2Fe+3H_2O\\ PbO+H_2 \to Pb+H_2O\\ O_2+2H_2 \to 2H_2O\)