Zn hay ZnO ?
nFe = 3/14
nZn = 4/65
Fe2O3 + 3H2 ➝ 2Fe + 3H2O
..............9/28......3/14
ZnO + H2 ➝ Zn + H2O
..........4/65.....4/65
SUY RA nH2 = 697/1820
VH2 \(\approx\)8,578 l
\(m_{Zn}=16-12=4\left(g\right)\)
PTHH: \(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\left(1\right)\)
\(ZnO+H_2\rightarrow Zn+H_2O\left(2\right)\)
\(n_{Fe}=\frac{12}{56}=0,21\left(mol\right)\)
\(n_{Zn}=\frac{4}{64}=0,0625\left(mol\right)\)
Theo PTHH (1): \(n_{Fe}:n_{H_2}=2:3\)
\(\Rightarrow n_{H_2\left(1\right)}=n_{Fe}.\frac{3}{2}=0,21.\frac{3}{2}=0,315\left(mol\right)\)
\(\Rightarrow V_{H_2\left(1\right)}=0,315.22,4=7,056\left(l\right)\)
Theo PTHH (2): \(n_{Zn}:n_{H_2\left(2\right)}=1:1\)
\(\Rightarrow n_{H_2\left(2\right)}=n_{Zn}=0,0625\left(mol\right)\)
\(\Rightarrow V_{H_2\left(2\right)}=0,0625.22,4=1,4\left(l\right)\)
Vậy thể tích khí H2 dùng để khử 2 hỗn hợp oxit trên là:
\(7,056+1,4=8,456\left(l\right)\)
