What mass of aluminum hydroxide is produced when 50 mL of .2 Mol Al(NO3)3 reacts with 200 mL of .1 mole KOH?

Question

What mass of aluminum hydroxide is produced when 50 mL of .2 Mol Al(NO3)3 reacts with 200 mL of .1 mole KOH?

Answer 1

Al(NO3)3 + 3KOH ----->  Al(OH)3 + 3KNO3

M = mol / L

mol = M x L = 0.200 mol/L x 0.0500 L = 0.0100 mol Al(NO3)3

mol = M x L = 0.100 mol/L x 0.200 L = 0.0200 mol KOH

Find the limiting reagent -

1 mol Al(NO3)3 reacts with 3 mol KOH

There are only 2 times as many moles KOH as there are Al(NO3)3, but 3 times as many is required. KOH is the limiting reagent. Use KOH to determine the yield.

3 mol KOH produces 1 mol Al(OH)3

0.0200 mol KOH x (1 mol Al(OH)3 / 3 mol KOH) = 0.00667 mol Al(OH)3

The molar mass of Al(OH)3 is 78.0 g/mol.

0.00667 mol Al(NO3)3 x (78.0 g / 1 mol) = 0.052 g Al(OH)3