PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(2Zn+O_2\underrightarrow{t^o}2ZnO\)
Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\) ⇒ 27x + 65y = 2,38 (1)
Ta có: \(n_{O_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{O_2}=\dfrac{3}{4}x+\dfrac{1}{2}y=0,04\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,04\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%n_{Al}=\dfrac{0,04}{0,04+0,02}.100\%\approx66,67\%\\\%n_{Zn}\approx33,33\%\end{matrix}\right.\)
mAl = 0,04.27 = 1,08 (g)
mZn = 0,02.65 = 1,3 (g)
