Ta có: \(n_H=\dfrac{0,12}{1}=0,12\left(mol\right)\)
Có: \(n_{C_2H_6O}=\dfrac{1}{6}n_H=0,02\left(mol\right)\)
\(\Rightarrow V_{C_2H_6O}=0,02.22,4=0,448\left(l\right)\)
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