What is the percent yeild of soduim sulfate when 32.18 g of sulfric acid reacts with excess sodium hydroxide to produce 37.91 g of sodiumsulfate

Answer 1

Hi Bea,

I assume you have the formula that % yield = actual yield/ theoretical yield. You are given the actual yield of 37.91g, so you only need to find the theoretical yield to find your answer.

Your first step would be to determine (or look up) the chemical equation for the reaction going on here. What you'll find is that one mole of sulfuric acid becomes one mole of sodium sulfate (IF there's an excess of sodium hydroxide, which you do have in this problem).

Once you know this ratio, you'll need to determine the molar mass of each molecule (only the 2 molecules we care about). So, based on the 1:1 ratio in the chemical equation, if you had the molar mass of sulfuric acid (in grams), it would yield the molar mass of sodium sulfate (in grams).

Now you can set up a ratio that relates the molar mass of each molecule to the numbers in your problem. So: (molar mass of sulfuric acid) / (molar mass of sodium sulfate) = 32.18g / ???

Solve the ratio to determine the theoretical yield of sodium sulfate.

Finally, go back to your percent yield equation, using the value you just calculated and the actual yield, which is given in the problem.

Good luck!