\(2Mg+O_2\rightarrow2MgO\)
2a_____ a_______2a
\(4Al+3O_2\rightarrow2Al_2O_3\)
4b____3b_____2b
\(n_{O2}=\frac{2,8}{22,4}=0,125\left(mol\right)\)
\(\Rightarrow a+3b=0,125\left(1\right)\)
Ta có:
\(m_{MgO}=m_{Al2O3}=2a.40+2b.102=80a+204b=9,1\left(2\right)\)
Từ (1) va (2), ta có:
\(\left\{{}\begin{matrix}a+b=0,125\\80a+204b=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,05\\b=0,025\end{matrix}\right.\)
\(\left\{{}\begin{matrix}m_{Mg}=24.2a=24.2.0,05=2,4\left(g\right)\\m_{Al}=27.4b=27.4.0,025=2,7\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m=m_{Mg}+m_{Al}=2,4+2,7=5,1\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\frac{2,4}{5,1}.100\%=47,06\%\\\%m_{Al}=100\%-47,06\%=52,94\%\end{matrix}\right.\)
{Mg; Al} + 0,125 mol O2 → 9,1 gam hỗn hợp oxit.
Bảo toàn khối lượng có: m = 9,1 – 0,125. 32 = 5,1 gam