Difficult Stoichiometry problem about mole ratios.

Given the equation Na2SO3 + 2HCl > 2NaCl + SO2 + H2O, what mass of SO2 can be made from 25.0 grams of Na2SO3 and 22.0 grams of HCl? I got12.7 grams of SO2, but the answer key says 11.3 grams of SO2, can someone explain what I did wrong?

This is a limiting reactant problem where you have to find out which one is the limiting factor.

While you can set up 2 separate stoichiometric calculations, and the one with the least/smallest result will be your answer.

I, however, set up a mole ratio and find the limiting reactant FIRST, that way I only have to do one calculation/equation.

1. First, find the moles of each.

Grams -> moles

Given gram x 1 mol/molar mass


Molar mass Na2SO3: 126.04 g/mol

Molar mass HCl: 36.46 g/mol


25.0 g Na2SO3 x 1 mol Na2SO3/126.04 g Na2SO3 = 0.198 mol Na2SO3

22.0 g HCl x 1 mol HCl/36.46 g HCl = 0.603 mol HCl


2. Then using the balanced equation, set up a mole ratio: Coefficients/Number of moles.

1 Na2SO3/0.198 mol Na2SO3 = 2 HCl/x

We pretend we don't know the moles of HCl.

By crossing multiplying, we find x = 0.396 mol HCl, which is less than 0.603 mol HCl. This tells us we will have EXTRA HCl, and that Na2SO3 is our limiting reactant.


3. Do a regular stoichiometry problem:

25.0 g Na2SO3 x 1 mol Na2SO3/126.04 g Na2SO3 x 1 mol SO2/1 mol Na2SO3 x 64.06 g SO2/1 mol SO2 = 12.71 g SO2

Sig figs... 12.7 g SO2


Hmm, I seems to be getting 12.7 g SO2 also. I went ahead and calculate the theoretical yield for 22.0 g HCl, which is 19.3 g SO2.

I'm 100% certain that it's an answer key typo.

Good job!

You have to take into account that one of these reactants will be a limiting reagent and work your problem starting with th emass of the other reactant that you didnt use the first time you calculated it.

Its either: 25/molar mass of Na2So3 X (1/1) x molar mass of SO2

or if its the other reactant, try: 22/37 x (1/2) x molar mass of SO2

one of these has to be correct