Solution Stoichiometry

What mass of solid aluminum Hydroxide can be produced when 50 mL of 0.200 M Al (NO3)3 is added to 200 mL of 0.100 M solution of AgNO3?

Help!!! I need detailed work I don't know what is going wrong
Step 1: Al(NO3)3 + 3 KOH ---> AlOH3 + 3KNO3

Step 2: Convert to mol: Al(NO3)3 --- 0.01 mol KOH 0.02 mol

Step 3: KOH is limiting because there should be 0.03 mol. Multiply 0.02 mol * 1/3 = 0.0067 mol of AlOH3. Convert to grams

Book answer is 0.520 g I get around 0.3 g What happened?

Easy fix!
you did all the calculations correctly
except
you used AlOH3 when it should have been Al(OH)3
this has molar mass of about 78
and
0.0067 mol * 78 g/mol = 0.52 g

Your molar mass of aluminum hydroxide. The correct formula for alumiinum hydroxide is Al(OH)3 which makes its molar mass out to be 78.01 g/mol.

0.00667 mol x 78.01 g/mol = 0.520 g

Congrats of getting that far. Just a simple formula error did you in.