Bài 1: \(Fe\left(0,2\right)+2HCl\left(0,4\right)\rightarrow FeCl_2+H_2\left(0,2\right)\)
\(C_{MddHCl}=\dfrac{0,4}{0,3}=1,3M\)
\(V_{H_2}=4,48l\)
Bài 2: \(2Al\left(0,2\right)+3H_2SO_4\left(0,3\right)\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(0,3\right)\)
\(C\%ddH_2SO_4=\dfrac{0,3.98.100}{250}=11,76\%\)
\(V_{H_2}=6,72l\)