Sửa đề: 11,155 (l) → 11,1555 (l)
Ta có: \(n_{H_2}=\dfrac{11,1555}{24,79}=0,45\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
____0,3_____0,45_________________0,45 (mol)
⇒ a = 0,3.27 = 8,1 (g)
\(C_{M_{H_2SO_4}}=\dfrac{0,45}{0,2}=2,25\left(M\right)\)