\(n_{H_2O}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(2H_2+O_2\underrightarrow{t^o}2H_2O\)
0,2 0,2
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\\
\%V_{H_2}=\dfrac{4,48}{4,48+6,72}.100\%=40\%\\
\Rightarrow\%V_{O_2}=100\%-40\%=60\%\)
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