Question

Cần dieu che 33.6 gam Fe Cần vua du V lit khí H2 de khu Fe3O4.TÍNH V, khoi luong Fe3O4 can dung ?

Answer 1

\(n_{Fe}=\dfrac{33.6}{56}=0.6\left(mol\right)\)

\(Fe_3O_4+4H_2\underrightarrow{^{t^0}}3Fe+4H_2O\)

\(0.2..........0.8........0.6\)

\(m_{Fe_3O_4}=0.2\cdot232=46.4\left(g\right)\)

\(V_{H_2}=0.2\cdot22.4=17.92\left(l\right)\)

Answer 2

n Fe = 33,6.56 = 0,6(mol)

$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$

Theo PTHH :

n H2 = 4/3 n Fe = 0,8(mol) => V H2 = 0,8.22,4 = 17,92 lít

n Fe3O4 = 1/3 n Fe = 0,2(mol) => m Fe3O4 = 0,2.232 = 46,4 gam