How many moles of al2(so4)3 can form under these conditions?

A. Since you are starting with equal moles of the two reactants, you can see from the balanced equation that you need 3 moles of H2SO4 for every 2 moles of Al(OH)3. So, H2SO4 is the limiting reactant. 

B. 0.420 mol H2SO4 X (1 mol Al2(SO4)3 / 3 mol H2SO4) = 0.140 mol Al2(SO4)3 

C. Moles of Al(OH)3 used:

0.420 mol H2SO4 X (2 mol Al(OH)3/3 mol H2SO4)
= 0.280 mol Al(OH)3 used Moles remaining
= 0.420 - 0.280
= 0.140 mol Al(OH)3 remaining.