Đáp án:
\[m_{Mg}=19,2\times 12,5\%=2,4(g)\]
\[\to m_{Fe}=19,2-2,4=16,8(g)\]
\[n_{Mg}=\dfrac{2,4}{24}=0,1(mol)\]
\[n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\]
Phản ứng xảy ra:
\[2Mg+O_2\to 2MgO\]
\[3Fe+2O_2\to Fe_3O_4\]
Theo PTHH:
$∑n_{O_2}=\dfrac{1}{2}.n_{Mg}+\dfrac{2}{3}.n_{Fe}=\dfrac{1}{2}.0,1+\dfrac{2}{3}.0,3=0,25(mol)$
$\to m_{O_2}=0,25.32=8(g)$