a. PTHH: \(2H_2+O_2\rightarrow2H_2O\)
b. \(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)
Theo PTHH: \(n_{H_2}=n_{O_2}\cdot\dfrac{2}{1}=0,1\cdot\dfrac{2}{1}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c. Chưa hiểu đề bài lắm :))
d. Ta có: \(n_{O_2}=0,1\left(mol\right);n_{H_2}=\dfrac{3,6}{22,4}\approx0,16\left(mol\right)\)
Do \(0,1< 0,16\) nên \(H_2\) dư \(0,16-0,1=0,06\left(mol\right)\)