1. For the balanced equation shown below, if the reaction of 40.8 grams of C6H6O3 produces a 39.0% yield, how many grams of H2O would be produced ?
2. For the balanced equation shown below, if the reaction of 20.7 grams of CaCO3 produces 6.81 grams of CaO, what is the percent yield?
Answer 1 : you have to find the actual yield of H2O in this problem and you have given the percentage yield, here how to solve this:
The balanced equation here provides us the info that 1 mole of C6H6O3 reacts to form 3 mole of H2O. So, if we convert mass in grams of C6H6O3 using this mole-mole ratio we can find the theoretical yeild of H2O
40.8 grams of C6H6O3 to mooles of C6H6O3 (using molar mass of C6H6O3)
40.8 / 126.11 = 0.33 moles
0.33 moles C6H6O3 x 3 moles H2O / 1 mole C6H6O3 = 0.99 moles of H2O
now convert moles of mass in grams of H2O (using molas mass of H2O)
0.99 x 18.0 g = 17.8 grams ( theoretical yield)
now the percent yield is :
actual yield/ theroretical yeild x 100% = 39.0%
actual yield / 17.8 x 100% = 39.0%
actual yield of H2O = 6.83 grams
Answer 2 : 58.7%