LIMITING REACTANTS PERCENT YIELD ACTUAL AND THEORETICAL YIELD

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LIMITING REACTANTS, PERCENT YIELD, ACTUAL AND THEORETICAL YIELD

LIMITING REACTANTS, PERCENT YIELD, ACTUAL AND THEORETICAL YIELD

LIMITING REACTANTS • The reactant that limits the amount of product that can form

LIMITING REACTANTS • The reactant that limits the amount of product that can form in a chemical reaction. • Excess reactant is not completely used in the reaction.

PRACTICE Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride

PRACTICE Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. Si. O 2(s) + 4 HF(g) → Si. F 4(g) + 2 H 2 O(l) If 6. 0 mol HF is added to 4. 5 mol Si. O 2, which is the limiting reactant?

SAMPLE PROBLEM F SOLUTION SIO 2(S) + 4 HF(G) → SIF 4(G) + 2

SAMPLE PROBLEM F SOLUTION SIO 2(S) + 4 HF(G) → SIF 4(G) + 2 H 2 O(L) • Given: HF 6. 0 mol Si. O 2 4. 5 mol • Unknown: limiting reactant • Mol HF X mol Si. F 4/mol HF = mol Si. F 4 produced • Mol Sio 2 X mol Si. F 4/mol Si. O 2 = mol Si. F 4 produced • What is the limiting reactant? • HF

• Theoretical yield- the max amount of product produced from given amount of

• Theoretical yield- the max amount of product produced from given amount of reactant. • Actual yield- product is measured amount of product obtained from reaction. • Percentage yield- ratio of actual yield to theoretical yield, multiplied by 100. • Percentage yield = actual yield/theoretical yield X 100

PERCENTAGE YIELD, CONTINUED Sample Problem Chlorobenzene, C 6 H 5 Cl, is used in

PERCENTAGE YIELD, CONTINUED Sample Problem Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C 6 H 6, with chlorine, as represented by the following equation. C 6 H 6 (l) + Cl 2(g) → C 6 H 5 Cl(l) + HCl(g) When 36. 8 g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38. 8 g. What is the percentage yield of C 6 H 5 Cl?

PERCENTAGE YIELD, CONTINUED Sample Problem Solution C 6 H 6 (l) + Cl 2(g)

PERCENTAGE YIELD, CONTINUED Sample Problem Solution C 6 H 6 (l) + Cl 2(g) → C 6 H 5 Cl(l) + HCl(g) Given: mass of C 6 H 6 = 36. 8 g mass of Cl 2 = excess actual yield of C 6 H 5 Cl = 38. 8 g Unknown: percentage yield of C 6 H 5 Cl Solution: Theoretical yield molar mass factor mol ratio molar mass

PERCENTAGE YIELD, CONTINUED Sample Problem H Solution, continued C 6 H 6(l) + Cl

PERCENTAGE YIELD, CONTINUED Sample Problem H Solution, continued C 6 H 6(l) + Cl 2(g) → C 6 H 5 Cl(l) + HCl(g) Theoretical yield Percentage yield

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Limiting Reactants and Percent Yield Definitions The LimitingLimiting Reactants and Percent Yield Definitions The Limiting
Limiting Reactants and Percent Yield Definitions The LimitingLimiting Reactants and Percent Yield Definitions The Limiting