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!
- 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
- LiOH + KClLiCl + KOH
- 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
- I actually produced 6.0 grams of lithium chloride. What is my percent yield?
6.0 x 100 = 16%
37
- C3H8 + 5O23CO2 + 4H2O
- 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
- 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
- Sodium chloride + Calcium oxide Calcium chloride + Sodium oxide
- Write the balanced chemical equation
2NaCl + CaO CaCl2 + Na2O
- 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
- 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%
- Given the following equation: ____K2CO3 + _2_HCl ------> ____H2O + ____CO2 + _2_KCl
- 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
- 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%
- 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%
- Given the following equation: _____ K2PtCl4+ _2__ NH3 ------> _____ Pt(NH3)2Cl2+ _2__ KCl
- 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
- 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%
- 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%
- 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%