When 46.0g of sodium hydroxide reacts with an excess of aluminum sulfate, how many grams of aluminum hydroxide are formed, according to the reaction below?

Question

1. When 46.0g of sodium hydroxide reacts with an excess of aluminum sulfate, how many grams of aluminum hydroxide are formed, according to the reaction below?

Al2(SO4)3(aq) + NaOH(aq) → Al(OH)3(s) + Na2SO4(aq)

1. When 46.0g of sodium hydroxide reacts with 465.3g of aluminum sulfate according to the reaction in the previous question (1), determine the limiting reagent and use it to find how many grams of aluminum hydroxide are expected to form?
2. If only 40.0g of aluminum hydroxide are formed, what is the percent yield of the reaction that took place in the previous question (2)?

Answer 1

Al₂(SO₄)₃(aq) + 6NaOH(aq) → 2Al(OH)₃(s) + 3Na₂SO₄(aq) .. balanced equation

molar mass NaOH = 39.997 g / mol

molar mass Al₂(SO₄)₃ = 342.15 g / mol

molar mass Al(OH)₃ = 78.00 g / mol

(1). 46.0 g NaOH x 1 mol NaOH / 39.997 g x 2 mol Al(OH)₃ / 6 mol NaOH x 78.00 g Al(OH)₃ / mol =

29.9 g Al(OH)₃

(1). To find the limiting reactant, we can divide the mols of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant:

NaOH: 46.0 g x 1 mol /39.997 g = 1.15 mols NaOH (÷6->0.19)

Al₂(SO₄)₃: 465.3 g x 1 mol / 342.15 g = 1.36 mols Al₂(SO₄)₃ (÷1->1.36)

Since 0.19 is less than 1.36, NaOH is the limiting reactant

Using moles of NaOH to find grams Al(OH)₃ formed:

1.15 mols NaOH x 2 mol Al(OH)₃ / 6 mols NaOH x 78.00 g / mol = 29.9 g Al(OH)₃ formed theoretically

(2). % yield = actual yield / theoretical yield (x100%0

% yield = 40.0 g / 29.0 g (x100%) = 134% yield. This is, of course, impossible. So, there is either a math error or an error in the question. My guess is that the question is incorrect as the actual yield must be equal to or less than 29.9 grams of Al(OH)₃