a) PTHH : \(2KCLO_3-t^0\rightarrow2KCL+3O_2\)
theo gt \(n_{KCLO3}=\frac{15,8}{122,5}\approx0,13\left(mol\right)\)
thep PTHH \(n_{O2}=\frac{3}{2}\cdot n_{KCLO3}=\frac{3}{2}\cdot0,13=0,195\left(mol\right)\\\Rightarrow V_{O2}=0,195\cdot22,4=4,368\left(l\right)\)
b) PTHH: \(3Fe+2O_2-t^0\rightarrow Fe_3O_4\)
theo gt: \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\), \(n_{O2}=0,195\left(mol\right)\)(theo câu a)
Theo PTHH: nFe =3(mol), nO2 =2(mol)
ta có tỉ lệ: \(\frac{0,2}{3}< \frac{0,195}{2}\Rightarrow\)O2 dư, tính số mol Fe3O4 theo Fe
ta có \(n_{Fe3O4}=\frac{1}{3}n_{Fe}=\frac{1}{3}\cdot0,2\approx0,067\left(mol\right)\)
\(\Rightarrow m_{Fe3O4}=0,067\cdot232=15,544\left(g\right)\)
2KmnO4 -nhiệt độ-> K2Mno4 + Mno2 + O2 (1)
nKmno4=m/M=0.1mol
theo pt (1) nO2=1/2nKmno4=0.05mol
=>vO2=n.22.4=0.05.22.4=1.12 lít
2O2 + 3Fe -nhiệt độ-> Fe3O4 (2)
0.05 0.2
0 0.125 0.05
=>mFe(dư)=n.M=7 gam
mFe3o4=n.M=11.6 gam