\(3Fe+2O_2\rightarrow Fe_3O_4\)
3 : 2
\(n_{Fe}=\frac{8.4}{56}=0.15\left(mol\right)\)
\(n_{O_2}=\frac{5.6}{22.4}=0.25\left(mol\right)\)
a)Vì \(\frac{0.15}{3}< \frac{0.25}{2}\)
\(\Rightarrow\) Oxi dư, bài toán tính theo Fe
\(n_{O_{2\left(pư\right)}}=\frac{2}{3}\cdot n_{Fe}=\frac{2}{3}\cdot0.15=0.1\left(mol\right)\)
\(\Rightarrow n_{O_{2\left(dư\right)}}=0.25-0.1=0.15\left(mol\right)\)
\(m_{O_{2\left(dư\right)}}=0.15\cdot32=4.8\left(g\right)\)
b) \(n_{Fe_3O_4}=\frac{1}{3}\cdot n_{Fe}=\frac{1}{3}\cdot0.15=0.05\left(mol\right)\)
\(m_{Fe_3O_4}=0.05\cdot232=11.6\left(g\right)\)
a) Ta có: nFe = 8,4/56 = 0,15 mol
nO2 = 5,6/22,4 = 0,25 mol => mO2 = 0,25.32 = 8g
PTHH: 3Fe + 2 O2 -> Fe3O4
3 mol 2 mol 1 mol
So sánh: 0,15/3mol < 0,25/2mol => Fe hết; O2 dư
Ta có: nO2 p/ứng : 0,15.2/3 = 0,1mol
=> mO2 p/ứng: 0,1 . 32 = 3,2 (g)
=> mO2 dư : 8 - 3,2= 4,8 (g)
b) Ta có: nFe3O4 = 0,15 . 1/3 = 0,05 (mol)
=> mFe3O4 = 0,05. 232 =11,6 (g)
