\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(4H_2+Fe_3O_4\underrightarrow{t^0}3Fe+4H_2O\)
4.............1...............3..........4 (mol)
0,4..........................0,3......... (mol)
\(m_{Fe\left(lithuyet\right)}=n.M=0,3.56=16,8\left(g\right)\)
\(m_{Fe\left(thucte\right)}=\dfrac{H\%.m_{Fe\left(lithuyet\right)}}{100}=\dfrac{92.16,8}{100}=15,456\left(g\right)\)
n\(_{H_2}\) = \(\dfrac{8,96}{22,4}\)= 0,4 (mol)
PTHH: Fe3O4 + 4H2 \(\underrightarrow{t^0}\) 3Fe + 4H2O
mol: --------------0,4---->0,3
m\(_{Fe}\) = 0,3 . 56 = 16,8 (g)
Vì H = 92% nên:
m\(_{Fe}\) = \(\dfrac{16,8.92}{100}\)= 15,456 (g)
