10.24 mL of 0.568 M Al(NO3)3 is mixed with 3.12 mL of 4.16 M NaOH. How much solid Al(OH)3 could be formed?
A) 0.055 g
B) 0.075 g
C) 0.111 g
D) 0.166 g
I tried finding how many moles of Al(OH)3 could be formed by using each of the two reactants to see which one was the limiting reagent. 0.01024 L * 0.568 M Al(NO3)3*1= 0.00581632 mol Al(OH)3 using the Al(NO3)3 and 0.00312 L* 4.16 M NaOH /3 = 0.0043264 mol Al(OH)3 using the NaOH. So the NaOH is the limiting reagent and there can be 0.0043264 mol Al(OH)3 * 78 g/mol = 0.337 g Al(OH)3 according to my calculations.
But this isn't any of the answer choices so am I doing anything wrong?