số mol Al là :
\(n_{Al}=\frac{2,7}{27}=0,1\left(mol\right)\)
PTHH:\(4Al+3O_2\rightarrow^t2Al_2O_3\)
\(V_{kk}=V_{O_2}:20\%=\frac{\left(n.22,4\right)}{20}.100=\frac{\left(0,1.\frac{3}{4}.22,4\right)}{20}.100=8,4\left(l\right)\)
\(m_{Al_2O_3}=n.M=\left(0,1.2:4\right).\left(27.2+16.3\right)=5,1\left(g\right)\)
số mol O2 là :
\(n=\frac{1,344}{22,4}=0,06\left(mol\right)\)
\(\frac{n_{O_2}đề}{n_{O_2}pt}=\frac{0,06}{3}=0,02< \frac{n_{Al}đề}{n_{Al}pt}=\frac{0,1}{4}=0,25\)
tính theo O2
\(m_{Al_2O_3}=n.M=\left(0,06.2:3\right).\left(27.2+16.3\right)=40,8\left(g\right)\)
