2X + O2 ----> 2XO (1)
=> \(m_O=20,4-14=6,4\left(g\right)\)
\(\Rightarrow n_O=\dfrac{6,4}{16}=32\left(mol\right)\)
XO +2HCl ----> XCl2 + H2O (2)
theo sơ đồ 2 :
\(n_{O_{\left(XO\right)}}=n_{O_{H2O}}=n_{H2O}=0,4\left(mol\right)\)
Mà \(n_{HCl}=2n_{H2O}=0,8\left(mol\right)\)
=> \(V_{HCl}=\dfrac{0,8}{1}=0,8\left(l\right)=800\left(ml\right)\)