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UNIT 1 TIER 6 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given Solve problems involving theoretical,

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Presentation on theme: "UNIT 1 TIER 6 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given Solve problems involving theoretical,"— Presentation transcript:

1 UNIT 1 TIER 6 Determine the limiting reactant and the reactant in excess when quantities of reacting substances are given Solve problems involving theoretical, experimental and percentage yield

2 Limiting Reactants The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The excess reactant is the substance that is not used up completely in a reaction.

3 Limited Reactants, continued Sample Problem F Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. SiO 2 (s) + 4HF(g) → SiF 4 (g) + 2H 2 O(l) If 6.0 mol HF is added to 4.5 mol SiO 2, which is the limiting reactant?

4 Limited Reactants, continued Sample Problem F Solution SiO 2 (s) + 4HF(g) → SiF 4 (g) + 2H 2 O(l) Given: amount of HF = 6.0 mol amount of SiO 2 = 4.5 mol Unknown: limiting reactant Solution:

5

6 Limited Reactants, continued Sample Problem F Solution, continued SiO 2 (s) + 4HF(g) → SiF 4 (g) + 2H 2 O(l)

7 Percentage Yield The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. The actual yield of a product is the measured amount of that product obtained from a reaction. The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.

8 Percentage Yield, continued Sample Problem H 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 Cl2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percentage yield of C 6 H 5 Cl?

9 Percentage Yield, continued Sample Problem H 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:

10 C 6 H 6 (l) + Cl 2 (g) → C 6 H 5 Cl(l) + HCl(g)

11 LAB CALCULATIONS Suppose that you used 10 ml of.1M KCl and reacted it with 20 ml of.1M Pb(C 2 H 3 O 2 ) 2. (a) find the limiting reactant (b) find the grams of precipitate (c) find the percentage yield

12 The limiting reactant determines the amount of the product that can be produced. So to find the mass of the PbCl 2 that can theoretically be produced: To find the percentage yield if.115g of PbCl 2 is actually produced:


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