What mass of solid aluminum hydroxide can be produced when 50.0mL of .2M Al(NO3)3 is added to 200mL .1M KOH?

Question

What mass of solid aluminum hydroxide can be produced when 50.0mL of .200M Al(NO3)3 is added to 200.0 mL .100M KOH?

Not sure how to do this..

Al(NO3)3 + 3KOH --> Al(OH)3 + 3KNO3

Answer 1

Al(NO3)3 + 3KOH --> Al(OH)3 + 3KNO3
onset: 0.1 mol 0.2 0 0
(0.05L*2M) (0.2L*1M)
rxn: -0.2/3 -0.2 +0.2/3 +0.2
end: 0.2/3 mol

Al(OH)3 = 0.2/3 mol * 78 g/mol = 5.2 gram

Answer 2

There are .05 x .2 = .01 moles of Al3+

and .1x.2 = .02 moles of OH-

3 moles of OH- take up 1 mole of Al3+

So, .02 mole of OH- will take up .02/3 mole of Al3+ = .00667 mole of Al3+

Thus mass = .00667x27+.02x17 = .52 g