2Mg + O2 \(\rightarrow\)2MgO (1)
3Fe + 2O2 \(\rightarrow\)Fe3O4 (2)
Đặt nMg=a
nFe=b
Ta có:
\(\left\{{}\begin{matrix}24a+56b=19,2\\40a+\dfrac{232}{3}b=27,2\end{matrix}\right.\)
a=0,1;b=0,3
mMg=24.0,1=2,4(g)
% Mg=\(\dfrac{2,4}{19,2}.100\%=12,5\%\)
% Fe=100-12,5=87,5%