PTHH: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Ta có: \(\dfrac{V_{H_2}}{2}< \dfrac{V_{O_2}}{1}\Rightarrow\dfrac{n_{H_2}}{2}< \dfrac{n_{O_2}}{1}\) => H2 hết, O2 dư
Theo PTHH: \(n_{H_2O}=n_{H_2}\Rightarrow V_{H_2O}=V_{H_2}=10\left(ml\right)\)
Theo PTHH: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}.n_{H_2}\Rightarrow V_{O_2\left(pư\right)}=\dfrac{1}{2}.V_{H_2}=5\left(ml\right)\)
=> \(V_{O_2\left(dư\right)}=10-5=5\left(ml\right)\)
=> \(V_{khí.sau.pư}=V_{H_2O}+V_{O_2\left(dư\right)}=10+5=15\left(ml\right)\)