\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....0,1....\dfrac{1}{15}.....\dfrac{1}{30}\\ b,V_{O_2}=\dfrac{1}{15}.22,4=\dfrac{112}{75}\left(l\right)\\ c,m_{Fe_3O_4}=\dfrac{1}{30}.232=\dfrac{116}{15}\left(g\right)\)
a/
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b/
Áp dụng công thức:
\(m=n.M=>n=\dfrac{m}{M}\)
\(=>n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}\)
\(n_{Fe}=0,1\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 2
0,1 x
\(=>x=0,1\cdot2:3=0,06=n_{O_2}\)
Áp dụng công thức
\(V=n.22,4=>V_{O_2}=n_{O_2}\cdot22,4\)
\(V_{O_2}=0,06\cdot22,4=1,344\left(l\right)\)
c/
\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
3 1
0,1 y
\(=>y=0,1\cdot1:3=0,03=n_{Fe_3O_4}\)
\(=>m_{Fe_3O_4}=n_{Fe_3O_4}.M_{Fe_3O_4}\)
\(m_{Fe_3O_4}=0,03\cdot232=6,96\left(g\right)\)
Vậy........
a) PTHH : \(3Fe+2O_2\rightarrow Fe_30_4\)
b) Ta có: \(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Theo PTHH ta có: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.0,1=0,07\left(mol\right)\)
=> \(V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,07.22,4=1,57\)( lít)
c) Theo PTHH ta có: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}.0,1=0,03\left(mol\right)\)
=> \(m_{Fe_3O_4}=n.M=0,03.232=6,96\left(g\right)\)
a, PTHH : 3Fe + 2O2 → Fe3O4
b, nFe= \(\dfrac{5,6}{56}\)= 0,1(mol)
Theo PTHH ta có : nO2= \(\dfrac{2}{3}\) * 0,1 = \(\dfrac{1}{15}\)(mol)
\(_{Vo2}\)= \(\dfrac{1}{15}\)*22,4= \(\dfrac{112}{75}\)(mol)
c, Theo PTHH ta có : n Fe3O4= \(\dfrac{1}{3}\)* 0,1= \(\dfrac{1}{30}\)(mol)
\(_{mFe3O4}\)= \(\dfrac{1}{30}\) * 232 = \(\dfrac{116}{15}\)(g)
Chúc bạn học tốt
