PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
2x 3x x 3x (mol)
Ta có: \(m_{ddsaup/ứ}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=54x+200-6x\left(g\right)\)
Dung dịch muối có nồng độ 10%
\(\Rightarrow\dfrac{342x}{54x+200-6x}=0,1\) \(\Rightarrow x=\dfrac{50}{843}\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=a\%=\dfrac{\dfrac{50}{243}\cdot98}{200}\cdot100\%\approx10,08\%\)