Question

For the following reaction, 3.45 grams of sulfuric acid are mixed with excess calcium hydroxide. The reaction yields 4.49 grams of calcium sulfate.


H2SO4 (aq) + Ca(OH)2 (s)= CaSO4 (s) + 2 H2O (l)





(1) What is the theoretical yield of calcium sulfate ?

grams



(2) What is the percent yield for this reaction ?

%

Answer 1

first find the limiting reagent
well the question tells us "excess CaOH" so are limiting reagent in not in excess meaning H2So4 is the limiting reagent.

molar mass H2So4 = 98.078 g mol

now find the moles
3.45g x mol/98.078g = .03517moles


our eqution is
1mol + 1mol <----->1mol + 2 mol
or same ratios
.03517mol + .03517mol<--->.03517mol + .7mol

so we have .03517 moles of CaS04 lets find out the grams

.03517molx(136.1406 g/mol) = 4.8 g
4.8g is our theoretical yield

the problem states that your actual yield is 4.49

so actual / theoretical x 100 = per cent yield

plugging them in

4.49/4.8 x 100==93.5% percent yield

pretty good yield!!!

anyway good luck keep practicing these types of problems

until you get proficient at them