\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^0}2Al_2O_3\)
\(0.4......0.3..........0.2\)
\(Al_2O_3:\) oxit bazo => Nhôm oxit
\(m_{Al_2O_3}=0.2\cdot102=20.4\left(g\right)\)
\(\rightarrow m_{Al_2O_3}=0,4.102=40,8\left(g\right)\)\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\)
-> Phân loại: Phản ứng hóa hợp
-> \(Al_2O_3\) : nhôm oxit
b) So sánh:
\(\dfrac{n_{Al}}{4}=0,1< \dfrac{n_{O_2}}{3}=0,1333\)
=> Al hết
\(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=0,2\left(mol\right)\)
