$1.A$

$n_{FeCl_2}=\dfrac{22.86}{127}=0.18(mol);n_{FeCl_3}=\dfrac{29.25}{56+106.5}=0.18(mol)$

Bảo toàn nguyên tố $Cl:$

$n_{HCl}=2n_{FeCl_2}+3n_{FeCl_3}=0.18  \times 2 + 0.18 \times 3=0.9(mol)$

Bảo toàn nguyên tố $H:$

$n_{H_2O}=\dfrac{n_{HCl}}{2}=0.45(mol)$

Bảo toàn khối lượng:

$m=22.86+29.25+18 \times 0.45-0.9 \times 36.5=27.36(gam)$

$2.C$

$n_{FeO}=n_{Fe_2O_3}$

`->`Quy hỗn hợp về $Fe_3O_4.$

$n_{hh}=\dfrac{9.28}{232}=0.04(mol)$

`->`$n_O=4n_{hh}=0.16(mol)$

$n_{HCl}=2n_{O}=2 \times 0.16=0.32(mol)$

`=>`$V=0.32(L)$

$3.A$

$n_{Ca(OH)_2}=\dfrac{35.52}{74}=0.48$

$n_{H_2}=0.2(mol)$

$n_{OH}=2n_{Ca(OH)_2}+n_{NaOH}=2 \times 0.48+\dfrac{a}{40}$ $(*)$

Bảo toàn $e:$

$n_{OH}=2n_{H_2}+2n_O$

`<=>`$2 \times 0.48+\dfrac{a}{40}=2 \times 0.2+2n_O$

`=>`$n_O=0.28+\dfrac{a}{80}$

Bảo toàn nguyên tố $Ca:$

$n_{Ca}=n_{Ca(OH)_2}=0.48(mol)$

Bảo toàn nguyên tố $Na:$

$n_{Na}=n_{NaOH}=\dfrac{a}{40}(mol)$

$m_{X}=m_{Ca}+m_{Na}+16n_{O}$

`<=>`$41.04=40 \times 0.48+23 \times \dfrac{a}{40}+16 \times(0.28+\dfrac{a}{80}) $

`=>`$a=22.4$

$4.D$

$n_{CO_3^{2-}}=n_{CO_2}=a(mol);n_{H_2}=b(mol)$

`->`$\begin{cases} a+b=0.5\\44a+2b=0.5(16.96 \times 2)=16.96 \end{cases}$

`=>`$\begin{cases} a=0.38\\b=0.12 \end{cases}$

$n_{KCl}=\dfrac{56.62}{74.5}=0.76(mol)$

$n_{MgCl_2}=\dfrac{0.571m}{95}(mol)$

Bảo toàn nguyên tố $K:$

$n_{K}=n_{KCl}=0.76(mol)$

Bảo toàn nguyên tố $Mg:$

$n_{Mg}=n_{MgCl_2}=\dfrac{0.571m}{95}(mol)$

Bảo toàn nguyên tố $Cl:$

$n_{Cl}=n_{KCl}+2n_{MgCl_2}=0.76+\dfrac{1.142m}{95}(mol)$

Mà $n_{Cl^-}=2n_{CO_3^{2-}}+2n_{H_2}+2n_O$

`=>`$n_O=\dfrac{0.571m}{95}-0.12 (mol)$

$m_X=m_{K}+m_{Mg}+60n_{CO_3^{2-}}+16n_O$

`<=>`$m=39 \times 0.76+24 \times \dfrac{0.571m}{95} +60 \times 0.38+16 \times (\dfrac{0.571m}{95}-0.12)$

`=>`$m=66.51$

$5.C$

$n_{NO}=0.18(mol)$

Quy hỗn hợp về :

$Fe_2O_3$ và $m=34.08+16 \times \dfrac{0.18 \times 3}{2}=38.4(gam)$

`=>`$n_{Fe_2O_3}=\dfrac{38.4}{160}=0.24(mol)$

Bảo toàn nguyên tố $Fe:$

$n_{Fe(NO_3)_3}=2n_{Fe_2O_3}=0.48(mol)$

`=>`$m_Y=0.48 \times 242=116.16(gam)$