\(n_{H_2}=\dfrac{6}{2}=3\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(2..............3.....................1.........3\)
\(m_{Al}=2\cdot27=54\left(g\right)\)
\(m_{H_2SO_4}=3\cdot98=294\left(g\right)\)
\(m_{Al_2\left(SO_4\right)_3}=1\cdot342=342\left(g\right)\)
a) $n_{H_2} = \dfrac{6}{2} = 3(mol)$
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
2............3.................1...............3..............(mol)
$m_{Al} = 2.27 = 54(gam)$
$m_{H_2SO_4} = 3.98 = 294(gam)$
b)
$m_{Al_2(SO_4)_3} = 1.342 = 342(gam)$