`n_NO={7,84}/{22,4}=0,35mol`

`3Cu+8HNO_3→3Cu(NO_3)2+2NO+4H_2O`

`x`        `{8x}/3`        `x`                    `{2x}/3`

`Fe+4HNO_3→Fe(NO_3)3+NO+2H_2O`

$⇒\left \{ {{64x+56y=23,6} \atop {\frac{2x}{3}+y=0,35}} \right.$

$x=0,15 ; y=0,25$

`%Cu={0,15.64.100}/{23,6}=40,7%`

`%Fe=59,3%`

`n_Cu(NO_3)2=0,15mol→m_Cu(NO_3)2=0,15.188=28,2g`

`n_Fe(NO_3)2=0,25mol→m_Fe(NO_3)2=0,25.242=60,5g`

`m_A=28,2+60,5=88,7g`

`n_HNO_3={8x}/3+4y={8}/3.0,15+4.0,25=1,4mol`

`V_HNO_3={1,4}/2.{110}/{100} =0,77l`

1. Đặt $n_{Cu}=x(mol);n_{Fe}=y(mol)$

$\Rightarrow 64x+56y=23,6(1)\\ n_{NO}=\dfrac{7,84}{22,4}=0,35(mol)\\ PTHH:\\ 3Cu+8HNO_3\to 3Cu(NO_3)_2+2NO\uparrow+4H_2O\\ Fe+4HNO_3\to Fe(NO_3)_3+NO\uparrow+2H_2O\\ \Rightarrow \dfrac{2}{3}x+y=0,35(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,15(mol)\\ y=0,25(mol) \end{cases}\Rightarrow \begin{cases} \%_{Cu}=\dfrac{0,15.64}{23,6}.100\%=40,68\%\\ \%_{Fe}=100\%-40,68\%=59,32\% \end{cases}\\ 2. n_{Cu(NO_3)_2}=x=0,15(mol);n_{Fe(NO_3)_3}=y=0,25(mol)\\ \Rightarrow m_{Cu(NO_3)_2}=0,15.188=28,2(g)\\ m_{Fe(NO_3)_3}=0,25.242=60,5(g)\\ \Rightarrow m_{\text{muối trong dd A}}=m_{Cu(NO_3)_2}+m_{Fe(NO_3)_3}=28,2+60,5=88,7(g)\\ 3,\Sigma n_{HNO_3}=\dfrac{8}{3}x+4y=1,4(mol)\\ \Rightarrow V_{dd_{HNO_3}}=\dfrac{1,4}{2}=0,7(l)\\ \Rightarrow V_{dd_{HNO_3}\text{(ban đầu)}}=\dfrac{0,7}{100\%-10\%}\approx 0,778(l)=778(ml)$