A

$n_{CO}$ = $x ( mol )$

$n_{$H_{2}$ }$ = $y ( mol )$

`⇒` $\frac{28x + 2y}{x + y}$ = $10,75 . 2 = 21,5$

`⇒` $6,5x = 19,5y$

`⇒` $%$$V_{CO}$ = $\frac{x}{x + y}$ . $100% = 75%$

`⇒` $%$$V_{$H_{2}$ }$ = $100 - 75 = 25$

B

$n_{Fe}$ = $\frac{16,8}{56}$ = $0,3 ( mol )$

$F_{e_{3} }$$O_{4}$ + $4CO$ `→` $3Fe$ + 4$CO_{2}$ 

$F_{e_{3} }$$O_{4}$ + $4H_{2}$ `→` $3Fe$ + $4H_{2}$$O$

Theo $pt$ : $n_{Fe3}$ $O_{4}$ = $\frac{1}{3}$ $n_{Fe}$ = $0,1 ( mol )$

`⇒` $m_{Fe3}$ $O_{4}$ = $0,1 . 232 = 23,2 (g)$

`⇒` $m = 23,2(g)$

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\(\begin{array}{l}
a)\\
M_B=10,75.2=21,5(g/mol)\\
\to \frac{n_{CO}}{n_{H_2}}=\frac{21,5-2}{28-21,5}=3\\
\to n_{CO}=3n_{H_2}\\
\to \%V_{CO}=\frac{3}{1+3}.100\%=75\%\\
\to \%V_{H_2}=100-75=25\%\\
b)\\
n_{Fe}=\frac{16,8}{56}=0,3(mol)\\
Fe_3O_4+4CO\xrightarrow{t^o}3Fe+4CO_2\\
Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O\\
Theo\,PT:\,n_{CO}+n_{H_2}=\frac{4}{3}n_{Fe}=0,4\\
Ma\,n_{CO}=3n_{H_2}\\
\to n_{CO}=0,3(mol);n_{H_2}=0,1(mol)\\
\to V=(0,3+0,1).22,4=8,96(l)\\
Theo\,PT:\,n_{Fe_3O_4}=\frac{1}{3}n_{Fe}=0,1(mol)\\
\to m=m_{Fe_3O_4}=0,1.232=23,2(g)
\end{array}\)