4Al+3O3--->2Al2O3
x----------------0,5x
2Mg+O2--->2MgO
y-------------y(mol)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}27x+24y=7,8\\51x+40y=14,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
m Al=0,2.27=5,4(g)
m Mg=0,1/24=2,4(g)
\(4Al+3O_2\underrightarrow{^{to}}2Al_2O_3\)
x___________0,5x
\(2Mg+O_2\underrightarrow{^{to}}2MgO\)
y_____________y
Gọi x, y lần lượt là nAl và nMg, ta có:
\(\left\{{}\begin{matrix}27x+24y=7,8\\102.0,5x+40y=14,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{Mg}=0,1.24=2,4\left(g\right)\)
