What mass of solid aluminum hydroxide can be produced when 50.0mL of 0.200M aluminum nitrate is added to 200.0mL of 0.100M potassium hydroxide

520 mg Al(OH)₃

Explanation:

The reaction that takes place is:

Now we calculate how many moles of each reagent were added, using their concentrations and volumes:

10 mmol of Al(NO₃)₂ would react completely with (10*3) 30 mmol of KOH, there are not as many KOH mmol, so KOH is the limiting reactant.

We calculate how many moles of Al(OH)₃ are produced, using the limiting reactant:

Finally we convert mmol of Al(OH)₃ to mg, using its molar mass:

answer: c.

explanation: