Ta có: \(n_{H_2}=\dfrac{0,028}{22,4}=0,00125\left(mol\right)\)
\(n_{O_2}=\dfrac{0,02}{22,4}=\dfrac{1}{1120}\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Xét tỉ lệ: \(\dfrac{0,00125}{2}< \dfrac{\dfrac{1}{1120}}{1}\), ta được O2 dư.
Theo PT: \(n_{H_2O}=n_{H_2}=0,00125\left(mol\right)\)
a, \(m_{H_2O}=0,00125.18=0,0225\left(g\right)\)
b, \(V_{H_2O}=0,00125.22,4=0,028\left(l\right)\)
Bạn tham khảo nhé!