a. \(n_{Fe}=\dfrac{1,8.10^{23}}{6.10^{23}}=0,3\left(mol\right)\)
\(n_{O_2}=\dfrac{6.72}{22,4}=0,3\left(mol\right)\)
PTHH : 3Fe + 2O2 -> Fe3O4
0,3 0,2 0,1
Ta thấy : \(\dfrac{0.3}{3}< \dfrac{0.3}{2}\) => Fe đủ , O2 dư
\(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
b. \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
\(V_{kk}=4,48.5=22,4\left(l\right)\)
a)\(n_{Fe}=\dfrac{1,8\cdot10^{23}}{6\cdot10^{23}}=0,3mol\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,3 0,3 0
0,3 0,2 0,1
0 0,1 0,1
\(m_{Fe_3O_4}=0,1\cdot232=23,2g\)
b)\(V_{O_2}=0,1\cdot22,4=2,24l\)
\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot2,24=11,2l\)
