Đặt $n_{Mg}=x(mol)\Rightarrow n_{Fe}=3x(mol)$

Mà $m_{Mg}+m_{Fe}=19,2$

$\Rightarrow 24x+56.3x=19,2$

$\Rightarrow x=0,1(mol)$

$\Rightarrow n_{Mg}=0,1(mol);n_{Fe}=0,3(mol)$

$a)PTHH:2Mg+O_2\xrightarrow{t^o}2MgO$

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

Từ 2 PT: $n_{MgO}=n_{Mg}=0,1(mol);n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1(mol)$

$\Rightarrow m_{oxit}=m_{MgO}+m_{Fe_3O_4}=0,1.40+0,1.232=27,2(g)$

$b)$ Từ 2 PT: $n_{O_2}=\dfrac{1}{2}.n_{Mg}+\dfrac{2}{3}n_{Fe}=\dfrac{1}{2}.0,1+\dfrac{2}{3}.0,3=0,25(mol)$

$\Rightarrow V_{O_2(đktc)}=0,25.22,4=5,6(lít)$

Đáp án:

\({m_{oxit}} = 27,2{\text{ gam}}\)

\({V_{{O_2}}} = 5,6{\text{ lít}}\)

Giải thích các bước giải:

 Gọi số mol \(Mg\) là \(x\); suy ra \(Fe\) là \(3x\)

\( \to {m_A} = {m_{Mg}} + {m_{F{\text{e}}}} = 24{\text{x}} + 56.3{\text{x}} = 192{\text{x}} = 19,2\)

\( \to x=0,1\)

Phản ứng xảy ra:

\(2Mg + {O_2}\xrightarrow{{{t^o}}}2MgO\)

\(3F{\text{e}} + 2{{\text{O}}_2}\xrightarrow{{{t^o}}}F{{\text{e}}_3}{O_4}\)

\( \to {n_{{O_2}}} = \frac{1}{2}{n_{Mg}} + \frac{2}{3}{n_{F{\text{e}}}} = \frac{1}{2}x + \frac{2}{3}.3{\text{x}} = 2,5{\text{x}} = 0,25{\text{ mol}}\)

Bảo toàn khối lượng:

\({m_A} + {m_{{O_2}}} = {m_{oxit}}\)

\( \to {m_{oxit}} = 19,2 + 0,25.32 = 27,2{\text{ gam}}\)

\({V_{{O_2}}} = 0,25.22,4 = 5,6{\text{ lít}}\)