0.4M Al(OH)3 reacts with 0.4M H2SO4 as follows:2Al(OH)3+3H2SO4→Al2(SO4)3+6H2OHow many moles of Al2(SO4)3 can form under these conditions?

Answer 1

0.4  X  1/3  =  0.1333333 mol Al2(SO4)3  << round for sig figs

the other reactant is excess

Answer 2

0.4mole of H2SO4 will react with 0.26mole of Al(OH)3.   0.26 mole of Al(OH)3 will give 0.13 mole of Al2(SO4)3.