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$2.D$ 

$n_{H_2SO_4}=0.025(mol)$

$M_Y=8.5 \times 4=34(amu)$

$n_{CO,dư}=a(mol);n_{CO_2}=b(mol)$

`->`$\dfrac{28a+44b}{a+b}=34$

`<=>`$28a+44b=34a+34b$

`<=>`$10b=6a$

`<=>`$a=\dfrac{10b}{6}=\dfrac{5b}{3}$

$\%V_{CO,dư}=\dfrac{\dfrac{5b}{3}}{\dfrac{5b}{3}+b} \times 100\%=62.5\%$

$\%V_{CO_2}=100\%-62.5\%=37.5\%$

$3.$

$n_{Fe}=\dfrac{4.368}{56}=0.078(mol)$

`->`$n_{e,max}=3n_{Fe}=0.234(mol)$

Bảo toàn khối lượng:

$m_{oxide}=m_{Fe}+m_{O_2}$

`=>`$n_{O_2}=\dfrac{6.096-4.368}{32}=0.054(mol)$

`=>`$n_O=2n_{O_2}=0.108(mol)$

$n_{e,chiếm}=2n_O=0.108 \times 2=0.216(mol)$

`->`$n_{e,còn}=n_{e,max}-n_{e,chiếm}=0.234-0.216=0.018(mol)$

$a)C$

Phần $1:$ $n_{e,chiếm}=\dfrac{0.216 }{3}=0.072(mol)$

Bảo toàn $e:$

$2n_{H_2}=n_{e,chiếm}$

`=>`$n_{H_2}=\dfrac{0.072}{2}=0.036(mol)$

`=>`$V_{H_2}=0.8064(L)$

$b)D$

Phần $2:$ $n_{e,còn}=\dfrac{0.018}{3}=0.006(mol)$

Bảo toàn $e:$

$3n_{NO}=n_e$

`=>`$n_{H_2}=\dfrac{0.006}{3}=0.002(mol)$

`=>`$V_{H_2}=0.0448(L)$

$c)D$

$n_{Al}=0.4(mol)$

Phần $3:$

$n_{O}=\dfrac{0.108}{3}=0.036(mol)$

$n_{Fe}=\dfrac{0.078}{3}=0.026(mol)$

Bảo toàn nguyên tố $O:$

$n_{Al_2O_3}=\dfrac{n_O}{3}=0.012(mol)$

Bảo toàn nguyên tố $Al:$

$n_{Al,pu}=2n_{Al_2O_3}=0.024(mol)$

`=>`$n_{Al,còn}=0.4-0.024=0.376(mol)$

Bảo toàn $e:$

$2n_{Fe}+3n_{Al,dư}=2n_{H_2}$

`=>`$n_{H_2}=\dfrac{2 \times 0.026+3 \times 0.376}{2}=0.59(mol)$

`=>`$V_{H_2}=13.216(L)$