$a)$ Bảo toàn KL: $m_{KL}+m_{O_2}=m_{oxit}$

$\Rightarrow m_{O_2}=17,4-11=6,4(g)$

$\Rightarrow n_{O_2}=\dfrac{6,4}{32}=0,2(mol)$

$\Rightarrow V_{O_2(đktc)}=0,2.22,4=4,48(lít)$

$b)$ Đặt $(n_{Fe};n_{Al})=(x;y)(mol)$

$\Rightarrow 56x+27y=11(1)$

$PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

Theo PT: $\dfrac{2}{3}x+\dfrac{3}{4}y=n_{O_2}=0,2(2)$

Từ $(1)(2)\Rightarrow x=\dfrac{19}{160}(mol);y=\dfrac{29}{180}(mol)$

$\Rightarrow m_{Fe}=\dfrac{19}{160}.56=6,65(g)$

$\Rightarrow \%m_{Fe}=\dfrac{6,65}{11}.100\%\approx 60,45\%$

$\Rightarrow \%m_{Al}=100-60,45=39,55\%$

$c)V_{O_2(đktc)}=4,48(lít)$

$\Rightarrow V_{kk(đktc)}=\dfrac{4,48}{20\%}=22,4(lít)$

Đáp án:

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