What is the percent yield of sodium sulfate (142.04 g/mol) from this reaction?

Question

Consider the reaction of 32.18 g of sulfuric acid (98.08 g/mol) with excess sodium hydroxide to produce 37.91 g of sodium sulfate. What is the percent yield of sodium sulfate (142.04 g/mol) from this reaction?

Answer 1

Hey Yasmine!

Percent yield is all about how much we actually create in a reaction, versus how much we expect to have produced if everything went ideally. In mathematical terms, percent yield is (actual yield)/(theoretical yield)x 100%. To find these values, we first must convert the mass of sulfuric acid to moles of sulfuric acid.

(Balanced) Reaction: H2SO4 + 2NaOH--> Na2SO4 + 2H2O (you can ignore the H2O, this was just to balance the equation and is excess information for the problem)

32.18g H2SO4 *1mole H2SO4/98.08 g H2SO4= 0.328 moles H2SO4

We would also expect that 0.328 moles of Na2SO4 to be created, because the ratio of Na2SO4 to H2SO4 is is 1:1. Converting this to grams we get:

0.328 moles Na2SO4 *142.04 grams/mole H2SO4 = 46.59 g Na2SO4.

46.59 g Na2SO4 is our theoretical yield, which means that if the reaction went to 100% completion and everything went perfectly, we could create 46.59 g Na2SO4 from the starting ingredients for this reaction. However, we were only able to create 37.91 g Na2SO4. This would be our actual yield. Going off the formula I mentioned earlier for percent yield:

Percent yield= (actual)/(theoretical)x 100%

= 37.91g/46.59g x 100%

Percent yield= 81.37%

Hope this helps!