a) 3Fe + 2O2 -to- Fe3O4
b) nFe = 0,3 (mol)
3Fe + 2O2 -to- Fe3O4
0,3.......0,2........0,1 (mol)
=> nO2 = 0,2.22,4 = 4,48 (lít)
c) Ta có phương trình:
2KClO3 -to- 2KClO2 + 3O2
215...................0,2215...................0,2 (mol)
=> mKClO3 = 215.122,5≈16,33
a) PTHH: \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
b)\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: \(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\left(1\right)\)
Theo PTHH: \(n_{O_2}=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) PTHH: \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\uparrow\left(2\right)\)
Theo PTHH: \(n_{KClO_3}=\dfrac{0,2.2}{3}=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow m_{KClO_3}=\dfrac{2}{15}.122,5=16,33\left(g\right)\)
3Fe + 2O2 -nhiệt độ-> Fe3o4 (1)
2Kclo3 -nhiệt độ-> 2kclo2 + O2 (2)
nFe=m/M=16.8/56=0.3 mol
nO2=2/3nFe=2/3.0.3=0.2mol
=> vO2=n.22.4=0.2.22.4=4.48 lít
theo pt (2) nkclo3=2nO2=0.4 mol
=>mkclo3=n.M=49 g
