\(n_{Fe}=\dfrac{m}{M}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
a) PTHH
3Fe + 2O2 --to--> Fe3O4
0,15.....0,1.................0,05 (mol)
=> \(m_{Fe_3O_4}=n\cdot M=0,05\cdot232=11,6\left(g\right)\)
b) \(\Rightarrow V_{O_2}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)
=> Vkhông khí = VO2 . 5 = 2,24 . 5 =11,2 (l)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Theo PT ta có: \(n_{Fe_3O_4}=\dfrac{1}{3}.n_{Fe}=\dfrac{1}{3}.0,15=0,05\left(mol\right)\)
a. Khối lượng oxit sắt thu được là:
\(m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\)
b. Theo PT ta có: \(n_{O_2}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
Mà oxi chiếm 1/5 thể tích kk
\(\Rightarrow V_{kk}=5.V_{O_2}=5.2,24=11,2\left(l\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
a) PTHH: \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
......3mol ......................1mol
...... 0,15mol................a mol
Gọi a là số mol của \(Fe_3O_4\)
Theo PTHH: \(\Rightarrow n_{Fe_3O_4}=a=0,05\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\)
b) Theo PTHH: \(\Rightarrow n_{O_2}=\dfrac{0,15.2}{3}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=5.2,24=11,2\left(l\right)\)