a) PTHH: \(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
Ta có tỉ lệ\(\dfrac{0,3}{2}< \dfrac{0,6}{2}\) => Al phản ứng hết, \(H_2SO_4\) dư
=> m\(H_2SO_4\left(dư\right)\) = \(0,6.98-\left(0,6-\dfrac{0,3.3}{2}\right).98=44,1\left(g\right)\)
c) \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\)
\(\Rightarrow V_{H_2\left(đktc\right)}=0,45.22,4=10,08\left(l\right)\)