How much sodium hydroxide must you start with to yield 145 g of aluminum hydroxide

Question

given the equation: Al₂(SO₃)₃+ 6NaOH ----> 3Na₂SO₃ + 2Al (OH)₃ and you start with 389.4 g of Al₂(SO₃)₃ and isolate 212.4 g of Na₂SO₃ How much sodium hydroxide must you start with to yield 145 g of aluminum hydroxide

Answer 1

Al₂(SO₃)₃+ 6NaOH ==> 3Na₂SO₃ + 2Al (OH)₃ ... balanced equation

From the stoichiometry of the balanced equation and using dimensional analysis, we can find how much NaOH you must start with.

389.4 g Al₂(SO₃)₃ x 1 mol/294 g x 3 mol Na₂SO₃/mol Al₂(SO₃)₃ x 126 g Na₂SO₃/mol = 500 g Na₂SO₃

% yield = 212.4 g/500 g (x100%) = 42.48%

145 g Al(OH)3 x 1 mol Al(OH)3/78 g x 6 mol NaOH/2 mol Al(OH)3 x 40 g NaOH/mol = 223 g NaOH

Correcting for % yield:

223 g / 0.4248 = 525 g NaOH