What mass of solid aluminum hydroxide is produced when 50.0 mL of 0.20 M aluminum nitrate is added to 200.0 mL of 0.100 M potassium hydroxide?

0.52g of Al(OH)₃

Explanation:

The reaction of aluminium nitrate, Al(NO₃)₃, reacts with KOH, as follows:

Al(NO₃)₃ + 3KOH → Al(OH)₃ + 3 KNO₃

That means 1 mole of aluminium nitrate reacts with 3 moles of potassium hydroxide.

To find moles of aluminium hydroxide we need to determine the limiting reactant finding moles of each reactant:

Moles aluminium nitrate:

0.050L * (0.20mol / L) = 0.01 moles Al(NO₃)₃

Moles KOH:

0.200L * (0.100mol / L) = 0.02 moles of KOH

Thus, for a complete reaction of 0.02 moles of KOH are needed (Using the chemical equation)

0.02 moles of KOH * (1 mol Al(NO₃)₃ / 3 mol KOH) = 0.0067 moles of Al(NO₃)₃.

As there are 0.01 moles of Al(NO₃)₃, this is the excess reactant and KOH is the limiting reactant.

The 0.02 moles of KOH produce:

0.02 moles of KOH * (1 mol Al(OH)₃ / 3 mol KOH) = 0.0067 moles of Al(OH)₃

In mass (Using molar mass of Al(OH)₃: 78g/mol):

0.0067 moles of Al(OH)₃ * (78g/mol) =