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Chemistry Lesson
Percent Yield
Transcript
Percent yield compares what actually formed to what the balanced equation says was possible.
Use actual yield divided by theoretical yield, then multiply by one hundred.
Balance the reaction first so the mole ratio between oxygen and potassium oxide is correct.
From 5.12 g of oxygen, the theoretical mass of K2O is 30.1 g.
With an actual mass of 20.40 g, the reaction finished at 67.8 percent yield.
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1. Balance the equation for potassium and oxygen.
2. 5.12 g O2 reacts with excess K. Find the theoretical K2O yield.
3. 20.40 g K2O is collected. Calculate the percent yield.
Stoichiometry & Percent Yield
1. Balance the reaction between potassium and oxygen to form potassium oxide.
2. If 5.12 g of O2 reacts with excess K, what theoretical mass of K2O should form?
3. A student actually collects 20.40 g of K2O. Calculate the percent yield.
4. Label which value is theoretical and which is actual.
Start with the reaction ratio. Balancing first tells us how oxygen and potassium oxide are related in moles.
Use the molar mass of oxygen gas so the mass on the worksheet becomes a usable mole quantity.
Apply the balanced-equation ratio, then convert moles of K2O into grams for the expected mass.
Once the worksheet has both values, percent yield is the actual mass divided by the theoretical mass times one hundred.
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