\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(n_{Al}=\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Ta thấy: \(\dfrac{0,1}{4}=\dfrac{1}{40}< \dfrac{1}{10}=\dfrac{0,3}{3}\)
do đó Al hết, O2 dư
b) Theo PTHH, \(n_{Al_2O_3}=\dfrac{2}{4}n_{Al}=0,5\cdot0,1=0,05\left(mol\right)\)
\(m_{Al_2O_3}=n\cdot M=0,05\cdot102=5,1\left(g\right)\)
\(n_{Al}=\dfrac{2,7}{27}0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 4Al + 3O2 \(\underrightarrow{to}\) 2Al2O3
Ban đầu: 0,1.........0,3.........................(mol)
Phản ứng: 0,1.......0,075.......................(mol)
Sau phản ứng: 0.......0,225.→ .....0,05.......(mol)
a) O2 dư, Al hết
b) \(m_{Al_2O_3}=0,05\times102=5,1\left(g\right)\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
Theo PT ta có: \(\dfrac{0,1}{4}=0,025< \dfrac{0,3}{3}=0,1\)
a. \(\Rightarrow\)Al hết. Còn \(O_2\) dư. Vậy ta tính theo \(n_{Al}\)
b. Theo PTHH ta có:
\(n_{Al_2O_3}=\dfrac{0,1.2}{4}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=n.M=0,05.102=5,1\left(g\right)\)
