Số mol :
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3mol\)
3\(H_2\) + \(Fe_2O_3\) --> 2\(Fe\) +3 \(H_2O\)
\(\Rightarrow n_{H_2}=\dfrac{0,3.3}{2}=0,45mol\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,45.22,4=10,08\)
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\(n_{Fe}\)=16,8:56=0,3(mol)
Ta có PTHH:
\(Fe_2O_3\)+3H2->2Fe+3H2O
.............0,45.....0,3............(mol)
Theo PTHH:\(n_{H_2}\)=0,45(mol)
=>\(V_{H_2\left(đktc\right)}\)=0,45.22,4=10,08(l)
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