a. PTHH: \(4Al+3O_{2}\) ➝ \(2Al_{2}O_{3}\)
Nhôm+Oxi➝ Nhôm Oxit
b+c. \(n_{Al_{đb}}=\frac{m_{Al}}{M_{Al}}=\frac{10,8}{27}=0,4\left(mol\right)\)
\(n_{O_{2_{đb}}}=\frac{V_{O_2}}{22,4}=\frac{4,032}{22,4}=0,18\left(mol\right)\)
\(4Al+3O_2\rightarrow2Al_2O_3\) (1)
Theo PTHH: 4 mol 3 mol
\(\frac{n_{Al_{đb}}}{n_{Al_{PTHH}}}\) \(\frac{n_{O_{2_{đb}}}}{n_{O_{2_{PTHH}}}}\)
\(\frac{0,4}{4}\) > \(\frac{0,18}{3}\)
\(\Rightarrow O_2\) hết; \(Al\) dư
Theo PTHH(1): \(n_{Al}=\frac{4}{3}n_{O_2}=\frac{4}{3}.0,18=0,24\left(mol\right)\)
\(\Rightarrow m_{Al\left(dư\right)}=0,24.27=6,48\left(g\right)\)
\(n_{Al_2O_3}=\frac{2}{3}n_{O_2}=\frac{2}{3}.0,18=0,12\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,12.102=12,24\left(g\right)\)
\(n_{Al}=\frac{10,8}{27}=0,4\left(mol\right)\)
\(n_{O2}=0,18\left(mol\right)\)
\(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)
___________4 _______3 ____ 2 (mol)
_______0,4_____0,18__________ (mol)
_______0,18 ___0,18___0,18 ( mol)
________0,22 __0 ______ 0_________ (mol)
b) Al là chất dư sau pư
\(m_{Al}=0,22.27=5,94\left(g\right)\)
c) \(m_{Al2O3}=0,18.102=18,36\left(g\right)\)
