\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{4}\) => H2 dư
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
0,1---->0,4------->0,3---->0,4
=> mFe = 0,3.56 = 16,8 (g)
mH2O = 0,4.18 = 7,2 (g)
mH2(dư) = (0,5-0,4).2 = 0,2 (g)
\(pthh:4H_2+Fe_3O_4\overset{t^o}{--->}3Fe+4H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\\n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\end{matrix}\right.\)
Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,5}{4}\)
Vậy H2 dư.
Theo pt: \(n_{Fe}=3.0,1=0,3\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
Theo pt: \(n_{H_{2_{PỨ}}}=4.0,1=0,4\left(mol\right)\)
\(\Rightarrow m_{H_{2_{dư}}}=\left(0,5-0,4\right).2=0,2\left(g\right)\)
molh2=11.2/22.4=0.5(mol)
molfe3o4=23.2÷232=0.1(mol)
PTHH:fe3o4+4h2--t°-->3fe+4h2o
mol----0.1----0.5
Ta thấy 0.1/1<0.5/4-->sau pư,fe3o4 pư hết,h2 dư.Tính theo fe3o4
-->sau pư thu đc fe,h2 dư,h2o
mol h2 pư=0.1×4÷1=0.4(mol)
-->mol h2 dư =0.5-0.4=0.1(mol)
-->mh2 dư=0.1×1=0.1(mol)
mfe =(0.1×3÷1)×56=16.8(g)
mh2o=(0.1×4÷1)×18=7.2(g)
