How many grams of Al(OH)3 (molar mass = 78.0 g/mol) can be produced from the reaction of 48.6 mL of .15 M KOH with excess Al2(SO4)3? Al2(SO4)3 + 6KOH - 2Al(OH)3 + 3K2SO4

Question

How many grams of Al(OH)3 (molar mass = 78.0 g/mol) can be produced

from the reaction of 48.6 mL of .15 M KOH with excess Al2(SO4)3?

Al2(SO4)3 + 6KOH - 2Al(OH)3 + 3K2SO4

Answer 1

Hello Maria,

in this reaction the limited reagent is KOH, so we have to base our calculations on it. The problem can be solved in three small parts:

1) Calculate the moles of KOH:

moles = molarity (M) * volume (L) = 0.15 (M) * (48.6 (mL) / 1000) = 0.15 (M) * 0.0486 (L) = 0.00729

so the moles of KOH in the reaction are 0.00729

2) Calculate the moles of Al(OH)3

we need 3 moles of KOH to form one mole of Al(OH)3 so, to calculate the maximum moles of Al(OH)3 that we can form we have to divide the moles of KOH by 3 :

moles of Al(OH)3 = moles of KOH / 3 = 0.0729 /3 = 0.00243

3) Calculate the grams of Al(OH)3

the problem gives us the molar mass of Al(OH)3 which is 78 grams per mole so:

g = moles * molar mass = 0.00243 moles * 78 grams/mole = 0.18954 grams

Hope this helps!