Name: ______ANSWER KEY______Date: ______Period: ______

Percent Yield Notes & Practice Worksheet

Theoretical yield: The maximum amount of product calculated using the balanced equation.

Actual yield: The amount of product obtained (what you actually got) when the reaction takes place.

Percent yield: The ratio of actual yield to theoretical yield.

percent yield = actual yield (g) x 100%

theoretical yield (g)

What is percent yield and why is it important?

When we perform stoichiometric calculations to determine the amount of product produced – that is a theoretical yield.

But when we actually perform the experiment in a lab setting, we usually find we do not get as much product, we usually get a smaller actual yield. (Human error, amirite?)

Practice Problems: Show all work, include units & proper significant figures. Don’t forget to balance first!

  1. Be + HCl BeCl2 + H2

My theoretical yield of beryllium chloride was 10.7 grams. If my actual yield was 4.5 grams, what was my percent yield?

Actual x 100 = 4.5 x 100 = 42%

Theoretical10.7

  1. LiOH + KClLiCl + KOH
  2. I began this reaction with 21 grams of lithium hydroxide. What is my theoretical yield of lithium chloride (in grams)?

21 g LiOH x 1 molLiOH x 1 molLiCl x 42.394 g LiCl = 37 g LiCl

23.9489 g LiOH 1 molLiOH 1 molLiCl

  1. I actually produced 6.0 grams of lithium chloride. What is my percent yield?

6.0 x 100 = 16%

37

  1. C3H8 + 5O23CO2 + 4H2O
  2. If I start with 5.30 grams of C3H8, what is my theoretical yield of water (in grams)?

5.30 g C3H8 x 1 mol C3H8 x 4 molH2O x 18.0158 g H2O = 8.66 g H2O

44.0952 g C3H8 1 mol C3H8 1 mol H2O

  1. I got a percent yield of 75.0%. How many grams of water did I actually make?

75% = actual x 1008.66 x 75 = actual x 8.66actual = 6.50 g H2O

8.66 100 8.66

  1. Sodium chloride + Calcium oxide  Calcium chloride + Sodium oxide
  2. Write the balanced chemical equation

2NaCl + CaO CaCl2 + Na2O

  1. What is my theoretical yield of sodium oxide if I start with 25 grams of calcium oxide?

25 g CaO x 1 molCaO x 1 molNa2O x 61.98 g Na2O = 28 g Na2O

56.078 g CaO 1 mol Na2O 1 mol Na2O

  1. H2SO4 H2O + SO3

If I start with 89 grams of sulfuric acid (H2SO4) and produce 7.10 grams of water, what is my percent yield?

89 g H2SO4 x 1 mol H2SO4 x 1 mol H2O x 18.0158 g H2O = 16 g H2O

98.0808 g H2SO4 1 mol H2SO4 1 mol H2O

7.10/16 x 100 = 44%

  1. Given the following equation: ____K2CO3 + _2_HCl ------> ____H2O + ____CO2 + _2_KCl
  2. Determine the theoretical yield of KCl if you start with 34.5 g of K2CO3.

34.5 g K2CO3 x 1 mol K2CO3 x 2molKCl x 74.551 g KCl = 37.2 g KCl

138.207 g K2CO3 1 mol K2CO3 1 molKCl

  1. Starting with 34.5 g ofK2CO3, and you isolate 3.4 g of H2O, what is the percent yield?

34.5 g K2CO3 x 1 mol K2CO3 x 1molH2O x 18.0158 g H2O = 4.5 g H2O

138.207 g K2CO3 1 mol K2CO3 1 molH2O

3.4/4.5 x 100 = 76%

  1. Given the following equation: ____H2SO4 + ____Ba(OH)2 ------> ____BaSO4 + _2_H2O

If 98.0 g of H2SO4 is reacted with excess Ba(OH)2, determine the percent yield of BaSO4 if you isolate 213.7 g of BaSO4.

98.0 g H2SO4 x 1 mol H2SO4 x 1 molBaSO4 x 233.395 g BaSO4 = 233 g BaSO4

98.0808 g H2SO4 1 mol H2SO4 1 molBaSO4

213.7/233 x 100 = 91.7%

  1. Given the following equation: _____ K2PtCl4+ _2__ NH3 ------> _____ Pt(NH3)2Cl2+ _2__ KCl
  2. Determine the theoretical yield of KCl if you start with 34.5 grams of NH3.

34.5 g NH3 x 1 mol NH3 x 2 molKCl x 74.551 g KCl = 151 g KCl

17.0307 g NH3 2 mol NH3 1 molKCl

  1. Starting with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the percent yield?

34.5 g NH3 x 1 mol NH3 x 1molPt(NH3)2Cl2 x 74.551 g Pt(NH3)2Cl2 = 75.5 g Pt(NH3)2Cl2

17.0307 g NH3 2 mol NH3 1 molPt(NH3)2Cl2

76.4/75.5 x 100 = 101%

  1. Given the following equation: ____H3PO4 + _3_KOH ------> ____K3PO4 + _3_H2O

If 49.0 g of H3PO4 is reacted with excess KOH, determine the percent yield of K3PO4 if you isolate 49.0 g of K3PO4.

49.0 g H3PO4 x 1 mol H3PO4 x 1 mol K3PO4 x 212.268 g K3PO4 = 106 g K3PO4

97.9977g H3PO4 1 mol H3PO4 1 mol K3PO4

49.0/106 x 100 = 46.2%

  1. Given the following equation: ____Al2(SO3)3 + _6_NaOH ------> _3_Na2SO3 +_2_Al(OH)3

If you start with 389.4 g of Al2(SO3)3 and you isolate 212.4 g of Na2SO3, what is your percent yield for this reaction?

389.4 g Al2(SO3)3 x 1mol Al2(SO3)3 x 3 mol Na2SO3 x 126.045 g Na2SO3 = 500.6 g Na2SO3

294.159 g Al2(SO3)3 1mol Al2(SO3)3 1 mol Na2SO3

212.4/500.6 x 100 = 42.43%