\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\\ n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,5}{4}>\dfrac{0,2}{2}\Rightarrow Aldư\\ \Rightarrow n_{O_2}=\dfrac{3}{2}.n_{Al_2O_3}=\dfrac{3.0,2}{2}=0,3\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)
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