\(2Cu+O_2\underrightarrow{^{to}}2CuO\)
a___________a
\(4Al+3O_2\rightarrow2Al_2O_3\)
a___________b/2
Giải hệ PT:
\(\left\{{}\begin{matrix}64a+27b=23,6\\80a+102.\frac{b}{2}=36,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\)
\(\%m_{Cu}=\frac{0,2.64}{23,6}.100\%=54,3\%\)
\(\%m_{Al}=100\%-54,3\%=45,7\%\)
Đặt \(n_{Cu}=x;n_{Al}=y\left(mol\right)\)
\(\Rightarrow64x+27y=23,6\left(1\right)\)
BTKL \(\Rightarrow m_{O_2}=36,4-23,6=12,8\left(g\right)\)
\(\Rightarrow n_{O_2}=0,4\left(mol\right)\)
QT nhường e: Cu ----> Cu2+ + 2e
_________ x_____________ 2x
Al ----> Al3+ + 3e
y_____________ 3y
QT nhận e: O2 + 4e ----> 2O2-
____________0,4___1,6
BT e \(\Rightarrow2x+3y=1,6\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)
