Question

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Answer 1

\(n_{Al}=\dfrac{10,8}{27}=0,4mol\)

4Al + 3O₂ \(\underrightarrow{t^o}\) 2Al₂O₃

0,4     0,3         0,2              ( mol )

\(\Rightarrow n_{Al_2O_3}=0,2mol\)

\(V_{O_2}=0,3.24,79=7,437l\)

Answer 2

a: \(4Al+3O_2\rightarrow2Al_2O_3\)(đk: t độ)

b: \(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)

nên \(n_{Al_2O_3}=0.2\left(mol\right)\)

\(m_{Al_2O_3}=0.2\cdot102=20.4\left(g\right)\)

Answer 3