Al 2 (SO 3 ) 3 + 6NaOH-- Na 2 SO 3 + 2Al(OH) 3

Question

Al 2 (SO 3 ) 3 + 6NaOH-- Na 2 SO 3 + 2Al(OH) 3

a) Calculate the limiting reactant if 12 g of Al 2 (SO 3 ) 3 is reacting with 12 g of NaOH.

b) Calculate the number of grams of Al(OH) 3 produced.

c) Calculate the number of grams of Na 2 SO 3 produced. (You can retrieve it form your

calculation in a.)

d) If 8 grams of Al 2 (SO 3 ) 3 were obtained, what is the percent yield?

Answer 1

First, you have to write a correctly balanced equation:

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

To find limiting reactant, we will compare moles of each reactant present considering their stoichiometric coefficients in the balanced equation.

moles Al2(SO3)3 present = 12 g Al2(SO3)3 x 1 mol/294 g = 0.0408 moles

moles NaOH present = 12 g NaOH x 1 mol/40 g = 0.30 moles

a). It might appear that NaOH is limiting but since it takes SIX moles NaOH for every 1 mol Al2(SO3)3, that now makes Al2(SO3)3 limiting (Just divide 0.30 by 6 and compare that to 0.0408 divided by 1)

b). grams Al(OH)3 produced: 0.0408 mol Al2(SO3)3 x 2 mol Al(OH)3 / mol Al2(SO3)3 x 78 g/mol = 6.4 g

c). grams Na2SO3 produced: 0.0408 mol Al2(SO3)3 x 3 mol Na2SO3 / mol Al2(SO3)3 x 126 g/mol = 15 g

d). % yield = 8 g/15 g (x100%) = 53% = 50% (1 sig. fig.)