1. Given the following unbalanced equation: Al(SO3)3 + NaOH
---> Na2SO3 + Al(OH)3

If you start with 389.4 grams of Al(SO3)3 and 200 grams of NaOH ?

Identify the limiting reactant:

Identify the reactant in excess and by how much:

How many grams of Na2SO, will be produced:

If you perform the same reaction but get 212.4 grams of Na2SO3, what is
your percent yield?

Al2(SO3)3 + 6NaOH = 3Na2SO3 + 2Al(OH)3

(a) moles Al2(SO3)3 = 10.0 g x 1 mol/294 g = 0.0340 moles Al2(SO3)3

moles NaOH = 10.0 g x 1 mol/40 g = 0.25 moles NaOH

Divide each by the coefficient in the balanced equation:

0.0340 Al2(SO3)3 /1 = 0.0340 moles

0.25 NaOH/6 = 0.0417 moles

Al2(SO3)3 is limiting as it is in the least supply

(b) 0.0340 moles Al2(SO3)3 x 2 moles Al(OH)3/mole Al2(SO3)3 = 0.0680 moles Al(OH)3 produced

(c) 0.0340 moles Al2(SO3)3 x 3 moles Na2SO3/mole Al2(SO3)3 = 0.102 moles Na2SO3

0.102 moles Na2SO3 x 126 g/mol = 12.9 g Na2SO3 produced

(d) Excess reagent is NaOH

moles NaOH used = 0.0340 moles Al2(SO3)3 x 6 moles NaOH/mole Na2(SO3)3 = 0.204 moles used

moles NaOH remaining = 0.25 - 0.204 = 0.046 moles NaOH remaining

grams NaOH remaining = 0.046 moles NaOH x 40 g/mole = 1.84 g NaOH left over

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