[LHM]

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?

[Borek]

What you are doing wrong? You are using faulty answer key.