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Understanding Stoichiometry and Percent Yield in Chemical Reactions

This guide explains the concepts of stoichiometry and percent yield, essential tools in chemistry for predicting product quantities from reactants. Learn about theoretical yield—the maximum product expected from a reaction—and actual yield, the quantity obtained in practice. Through examples, discover how to calculate percent yield using the formula: (Actual Yield x 100) / Theoretical Yield, indicating the efficiency of your reaction. Gain insight into how to apply these concepts with practical problems involving magnesium and hydrogen gas.

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Understanding Stoichiometry and Percent Yield in Chemical Reactions

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  1. April 7, 2014 • Today: Stoichiometry and % Yield

  2. Percent Yield Remember, stoichiometry is used to tell you how much product you can form from X amount of product without doing the reaction Percent yield tells you how much product you actually got in the lab compared to how much you could have got

  3. Theoretical yield • The maximum amount of product that can be formed from a given amount of reactant. • This is a value you calculate on paper • In other words: (Write your own definition of Theoretical yield here)

  4. Actual yield • The measured amount of a product obtained from a reaction • This is a measurement of a product formed in an actual chemical reaction • In other words: (Write your own definition of Actual yield here)

  5. Percent Yield Formula Actual yield x 100 = Theoretical yield

  6. Example 1 Mg(s) + 2H2O(g)  Mg(OH)2(s) + H2(g) If 16.2 g Mg are heated with excess H2O how many grams of hydrogen gas could theoretically be formed? Known: mass Mg Unknown: mass H2 Plan: g Mg  mol Mg  mol H2  g H2 Relationships: molar mass Mg = 24.31 g/mol, molar mass H2 = 2.02 g/mol, mole ratio = 1 mol Mg : 1 mol H2 16.2 g Mg | 1 mol Mg | 1 mol H2 | 2.02 g H2 = 16.2 x 2.02 = 1.35 g 24.31 g 1 mol Mg 1 mol H2 24.31

  7. Example 1 • B. If only 0.905 g H2 are actually formed, what is the percent yield of this reaction? Actual yield x 100 = % yield Theoretical yield 0.905 g H2 x 100 = 67.0% 1.35 g H2 (This means that you produced or collected only 67% of the product that it was possible to form with the amount of reactant you started with.)

  8. Example 2 CO(g) + 2H2(g)  CH3OH(l) If 11.0 g H2reacts with CO to produce 68.4 g CH3OH, what is the percentage yield of CH3OH? We need to determine the theoretical yield. Known: mass H2 Unknown: mass CH3OH Plan: g H2 mol Mg  mol H2  g H2 Relationships: molar mass H2= 2.02g/mol molar mass CH3OH= 32.04 g/mol mole ratio = 2molH2 : 1 molCH3OH

  9. Example 2 CO(g) + 2H2(g)  CH3OH(l) If 11.0 g H2reacts with CO to produce 68.4 g CH3OH, what is the percentage yield of CH3OH? Known: mass H2 Unknown: mass CH3OH Plan: g H2 mol Mg  mol H2  g H2 Relationships: molar mass H2 = 2.02 g/mol, molar mass CH3OH = 32.04 g/mol, mole ratio = 2mol H2 : 1 molCH3OH 11.0 g H2 | 1 molH2| 1 molCH3OH| 32.04 g CH3OH= 11.0 x 32.04 = 87.2 2.02 H2 2 molH2 1 molCH3OH 2.02 x 2 g

  10. Example 2 CO(g) + 2H2(g)  CH3OH(l) If 11.0 g H2reacts with CO to produce 68.4 g CH3OH, what is the percentage yield of CH3OH? Actual yield x 100 = % yield Theoretical yield 68.4 g CH3OH x 100 = 78.4% 87.2 g CH3OH

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