Answer:
0.52g of Al(OH)₃
Step-by-step 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) =
0.52g of Al(OH)₃