Al(NO3)3 + 3KOH -------> 3KNO3 + Al(OH)3
50 ml * .2 moles/ liter = .01 Moles of Al(NO3)3
200 ml * .1 moles/liter = .02 Moles of KOH
Since the ratio between the two reactants according to the chemical equation is 1:3, we would need .03 moles of one to fully react with .01 moles of the other. Since we don't, only 1/150 mole of the first reactant will react with the .02 moles of the second reactant. This will produce .02 moles of KNO3 as well as .01 moles of Al(OH)3
.02 moles KNO3 = .02(48 grams + 14 grams + 40 grams) = .02(102 grams) = 2.04 grams
