chem help please?

Good Man!!! You need the BALANCED formula for this calculation.
Notice the molar ratios are 2:3::1:6
First calculate the moles of the two reactants using the equation.
moles= mass(g) / Mr
mol(Al(OH)3) = 2.531g / ( 27 + (3(16 + 1))
mol(Al(OH)3) = 2.531 / 78 = 0.03244 moles (Equivalent to '2' in molar ratios)
mol(H2SO4) = 3.892 / ((2 x 1) + 32 + (4 x 16))
mol(H2SO4) = 3.892 / 98 = 0.033971 (Equivalent to '3' in molar ratios).

So the ratios should be 2:3:: 0.03244: 0.03397 , But they are NOT . So we must find the limiting reactant.
To give 100% product the molar ratios should be 2:3::0.03244 : 0.04866
Since the moles (H2SO4) is less than 0.04866 then H2SO4 is the limiting reactant.

So 0.03397 moles(H2SO4) is equivalent to '1'
Hence moles(Al2(SO4)3) = 0.03397 /3 = 0.01132 is produced.
Mr (Al2(SO4)3 = (27x2) +(32 +(4 x 16)3) = 54 + 156 = 210
Hence
Mass (Al2(O4)3) = 0.01132 x 210 = 2.38 g

Since

According to Wikipedia :
Molar mass of Al(OH)₃ = 78.00 g/mol
Molar mass of H₂SO₄ = 98.08 g/mol
Molar mass of Al₂(SO₄)₃ = 342.15 g/mol

Initial number of moles of Al(OH)₃ = (2.531 g) / (78.00 g/mol) = 0.03245 mol
Initial number of moles of H₂SO₄ = (3.892 g) / (98.08 g/mol) = 0.03968 mol

Balanced equation for the reaction :
2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
Mole ratio Al(OH)₃ : H₂SO₄ : Al₂(SO₄)₃ = 2 : 3 : 1

When 0.03968 mol H₂SO₄ completely reacts :
Number of moles of Al(OH)₃ needed = (0.03968 mol) × (2/3) = 0.02645 mol < 0.03245 mol
Al(OH)₃ is in excess. Then, H₂SO₄ completely reacts (the limiting reactant).

In the reaction :
Number of moles of H₂SO₄ reacted = 0.03968 mol
Number of moles of Al₂(SO₄)₃ formed = (0.03968 mol) × (1/3) = 0.01323 mol
Mass of Al₂(SO₄)₃ formed = (0.01323 mol) × (342.15 g/mol) = 4.527 g