\(3Fe+2O_2\rightarrow Fe_3O_4\)
0,06__0,04____0,02
\(n_{Fe3O4}=\frac{4,46}{56.3+16.4}=0,02\left(mol\right)\)
\(m_{Fe}=0,06.56=3,36\left(g\right)\)
\(m_{O2}=0,04.32=1,28\left(g\right)\)
\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
0,08 ___________________________0,04
\(\rightarrow m_{KMnO_4}=0,08.\left(39+55+16.4\right)=12,64\left(g\right)\)
a, PTHH ( I ) : \(3Fe+2O_2\rightarrow Fe_3O_4\)
b, \(n_{Fe_3O_4}=\frac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\frac{4,46}{56.3+16.4}\approx0,02\left(mol\right)\)
- Theo PTHH ( I ) : \(n_{Fe}=3n_{Fe_3O_4}=3.0,2=0,6\left(mol\right)\)
-> \(m_{Fe}=n_{Fe}.M_{Fe}=0,6.56=33.6\left(g\right)\)
- Theo PTHH ( I ) : \(n_{O_2}=2n_{Fe_3O_4}=2.0,2=0,4\left(mol\right)\)
-> \(m_{O_2}=n_{O_2}.M_{O_2}=0,4.32=12,8\left(g\right)\)
c, PTHH ( II ) : \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
- Theo PTHH ( II ) : \(n_{KMnO_4}=2n_{O_2}=2.0,4=0,8\left(mol\right)\)
-> \(m_{KMnO_4}=n.M=0,8.158=126,4\left(g\right)\)
Đáp án:
3Fe + 2O2 --> Fe3O4
0,06 0,04 0,02
nFe3O4=4,46/(56.3+16.4)=0,02
mFe=0,06.56=3,36g
mO2=0,04.32=1,28g
2KMnO4 --> K2MnO4 + MnO2+O2
0,08 0,04
mKMnO4=0,08.(39+55+16.4)=12,64g
