PTHH: Mg+H2SO4-----> MgSO4+H2
\(n_{Mg}=\dfrac{4,8}{24}=0,2\) mol
\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\) mol
Ta có tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,15}{1}\)
----> Tính theo H2SO4
Theo pt: \(n_{H_2}=n_{H_2SO_4}=0,15\) mol
=> VH2= 0,15.22,4= 3,36 l