UNIT 1TIER 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

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.

Limited Reactants, continued

Sample Problem FSilicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to thefollowing equation.

SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

If 6.0 mol HF is added to 4.5 mol SiO2, which is the

limiting reactant?

Limited Reactants, continued

Sample Problem F SolutionSiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

Given: amount of HF = 6.0 mol amount of SiO2 = 4.5 mol

Unknown: limiting reactantSolution:

4

2 42

mol SiFmol SiO mol SiF produced

mol SiO

Limited Reactants, continuedSample Problem F Solution, continued

SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l)

4

4

1mol SiF6.0 mol HF 1.5 mol SiF produced

4 mol HF

4

2 42

1mol SiF4.5 mol SiO 4.5 mol SiF produced

1mol SiO

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.

actual yieldpercentage yield 100

theorectical yield

Percentage Yield, continued

Sample Problem HChlorobenzene, C6H5Cl, 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, C6H6, with chlorine, as represented by the following equation.

C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)

When 36.8 g C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g.What is the percentage yield of C6H5Cl?

Percentage Yield, continuedSample Problem H Solution

C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)

Given: mass of C6H6 = 36.8 gmass of Cl2 = excessactual yield of C6H5Cl = 38.8 g

Unknown: percentage yield of C6H5ClSolution:

C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g)

LAB CALCULATIONS

Suppose that you used 10 ml of .1M KCl and reacted it with 20 ml of .1M Pb(C2H3O2)2. (a) find the limiting reactant (b) find the grams of precipitate (c) find the percentage yield

The limiting reactant determines the amount of the product that can be produced . So to find the mass of the PbCl2 that can theoretically be produced:

To find the percentage yield if .115g of PbCl2 is actually produced: