Gọi $n$ là hoá trị của kim loại M

$n_{NO} = \dfrac{4,48}{22,4} = 0,2(mol)$
Bảo toàn electron : $n.n_M = 3n_{NO} \Rightarrow n_M = \dfrac{0,6}{n}(mol)$
$\Rightarrow \dfrac{0,6}{n}.M = 19,2$
$\Rightarrow M = 32n$

Với n = 2 thì M = 64(Cu)

$n_{CuO} = n_{Cu} = 0,3(mol)$
$\Rightarrow m = 0,3.80 = 24(gam)$

$n_{Cu} = \dfrac{57,6}{64} = 0,9(mol)$
$Cu^0 \to Cu^{+2} + 2e$
$O_2 + 4e \to 2O^{2-}$

Bảo toàn electron  :$2n_{Cu} = 4n_{O_2}$
$\Rightarrow n_{O_2} = 1,8(mol)$
$V_{O_2} = 1,8.22,4 = 40,32(lít)$

A ơi a xem lại câu này đc k

$n_{NH_4NO_3} = \dfrac{8}{80} = 0,1(mol)$

$n_{Zn(NO_3)_2} = \dfrac{113,4}{189}=0,6(mol)$
$Zn^0 \to Zn^{2+} + 2e$
$N^{+5} + 8e \to N^{-3}$

Bảo toàn electron : $2n_{Zn} = 8n_{NH_4NO_3}$
$\Rightarrow n_{Zn} = 0,4(mol)$
mà : $n_{Zn(NO_3)_2} = n_{Zn} + n_{ZnO}$
$\Rightarrow n_{ZnO} = 0,6 - 0,4 = 0,2(mol)$
$m_{ZnO} = 0,2.81 = 16,2(gam)$

$m_{dd\ HNO_3} = 150.1,367 = 205,05(gam)$
$n_{HNO_3} = \dfrac{205,05.60\%}{63} = 1,952(mol)$

$n_S = \dfrac{6,4}{32} = 0,2(mol)$
$S + 6HNO_3 \to H_2SO_4 + 6NO_2 + 2H_2O$

Ta thấy : 

$n_S : 1 < n_{HNO_3} :6$ nên $HNO_3$ dư

$n_{NO_2} = 6n_S = 1,2(mol)$
$m_{NO_2} = 1,2.46 = 55,2(gam)$

$n_{N_2O} = \dfrac{560}{1000.22,4} = 0,025(mol)$

Gọi $n_{Mg} = a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b = 1,86(1)$
$Mg^0 \to Mg^{+2} + 2e$
$Al^0 \to Al^{+3} + 3e$
$2N^{+5} + 8e \to 2N^{+1}$

Bảo toàn electron : $2n_{Mg} + 3n_{Al} = 8n_{N_2O}$

$\Rightarrow 2a + 3b = 0,025.8(2)$

Từ (1)(2) suy ra : a = 0,01 ; b = 0,06

$m_{Mg} = 0,01.24 = 0,24(gam)$

$n_{Cu} = \dfrac{8}{64} = 0,125(mol)$
$n_{H^+} = n_{HNO_3} + 2n_{H_2SO_4} = 0,24(mol)$
$n_{NO_3^-} = n_{HNO_3} = 0,12(mol)$
$3Cu + 8H^+ + 2NO_3^- \to 3Cu^{2+} + 2NO + 4H_2O$
Ta thấy : 

$\dfrac{n_{NO_3^-}}{2} > \dfrac{n_{Cu}}{3} > \dfrac{n_{H^+}}{8} $

Do đó, $Cu,NO_3^-$ dư

$n_{NO} = \dfrac{1}{4}n_{H^+} = 0,06(mol)$
$V = 0,06.22,4 = 1,344(lít)$

Sau phản ứng, dung dịch A gồm : 

$Fe^{3+} : 0,12(mol)$
$Cu^{2+} : 2x(mol)$
$SO_4^{2-} : 0,12.2 + x = 0,24 + x(mol)$
Bảo toàn điện tích : $3n_{Fe^{3+}} + 2n_{Cu^{2+}} = 2n_{SO_4^{2-}}$
$\Rightarrow 0,12.3 + 2x.2 = 2(0,24 + x)$
$\Rightarrow x = 0,06(mol)$

Gọi $n_{NO} = a(mol) ; n_{NO_2} = b(mol)$
$n_{hh} = a + b = \dfrac{8,96}{22,4} = 0,4(mol)$
$M_{hh} = \dfrac{30a + 46b}{a + b } = 16,75.2 = 33,5(g/mol)$
$\Rightarrow a = 0,3125 ; b = 0,0875

$Al^0 \to Al^{+3} + 3e$
$N^{+5} + 3e \to N^{+2}$
$N^{+5} + 1e \to N^{+4}$
Bảo toàn electron : $3n_{Al} = 3n_{NO} + n_{NO_2}$
$\Rightarrow n_{Al} = \dfrac{41}{120}$
$\Rightarrow m = \dfrac{41}{120}.27 = 9,225(gam)$

\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO\uparrow+2H_2O\)

\(Al+6HNO_3\rightarrow Al\left(NO_3\right)_3+3NO_2+3H_2O\)

Gọi \(\left\{{}\begin{matrix}n_{NO}=x\\n_{NO_2}=y\end{matrix}\right.\) ( mol )

\(n_{hhk}=x+y=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(1\right)\)

\(\dfrac{M_{hhk}}{M_{H_2}}=16,75\Rightarrow M_{hhk}=16,75.2=33,5\)

\(\Leftrightarrow\dfrac{30x+46y}{x+y}=33,5\)

\(\Leftrightarrow30x+46y=13,4\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,3125\\y=0,0875\end{matrix}\right.\)

Theo pt: \(n_{Al}=x+\dfrac{y}{3}=0,3125+\dfrac{0,0875}{3}=\dfrac{41}{120}\)

\(\Rightarrow m_{Al}=27.\dfrac{41}{120}=9,225\left(g\right)\)

$n_{Cu} = \dfrac{12,8}{64} = 0,2(mol)$

Gọi $n_{NO} = a(mol) ; n_{NO_2} = b(mol)$
$\Rightarrow M_A = \dfrac{30a + 46b}{a + b} = 19.2 = 38(1)$

$Cu^0 \to Cu^{+2} + 2e$
$N^{+5} + 3e \to N^{+2}$
$N^{+5} + 1e \to N^{+4}$

Bảo toàn electron : $2n_{Cu} = 3n_{NO} + n_{NO_2}$
$\Rightarrow 0,2.2 = 3a + b(2)$

Từ (1)(2) suy ra : a = 0,1 ; b = 0,1

$V = (0,1 + 0,1).22,4 = 4,48(lít)$

Gọi $n_{NO} = a(mol) ; n_{NO_2} = 2a(mol) ; n_{N_2} = 2a(mol)$

Ta có : $n_{hh\ khí} = a + 2a + 2a = \dfrac{44,8}{22,4} = 2(mol)$
$\Rightarrow a = 0,4(mol)$

$Al^0 \to Al^{+3} + 3e$
$N^{+5} + 3e \to N^{+2}$
$N^{+5} + 1e \to N^{+4}$
$2N^{+5} + 10 \to 2N^0$

Bảo toàn electron : $3n_{Al} = 3n_{NO} + n_{NO_2} + 10n_{N_2}=10$
$\Rightarrow n_{Al} = \dfrac{10}{3}(mol)$
$\Rightarrow a = \dfrac{10}{3}.27 = 90(gam)$