Mar 11--Excess and Limiting Reactants Percent Yield

Imagine yourself making cheese sandwiches.

Excess Quantity: ie. pressure, temperature, concentration, etc. of a chemical reaction which don’t show in the balanced equation.

Since it is not possible for every atom or molecules of the reactants to come together in a chemical reaction, it is necessary sometimes to add more of one reactant than the equation predicts.

ie. One reactant is the Excess Quantity and some of it will be left over. The second reactant is used up completely, and is the Limiting Factor.

In Chemical Reactions, we calculate the amount of product formed when one of the reactants is excess.

Ex. How many grams of S is produced when 15.6 g of KMnO₄, 7.95 g of H₂S, and 15.3 g of H₂SO₄ are reacted?

Step #1--Balance the equation.

2 KMnO₄ + 5 H₂S + 3 H₂SO₄ --> K₂SO₄ + 2 MnSO₄ + 8 H₂O + 5 S

Step #2--Convert all reactants to the desired product and the smallest amount of product will actually be produced.

15.6 g KMnO₄ × (1 mol KMnO₄ / 158.0 g) ×(5 mol S / 2 mol KMnO₄) × (32.1g / 1 mol S) = 7.92 g S

7.95 g H₂S × (1 mol H₂S / 34.1 g) × (5 mol S / 5 mol H₂S) × (32.1g / 1 mol S) = 7.48 g S

15.3 g of H₂SO₄ × (1 mol H₂SO₄ / 98.1 g) × (5 mol S / 3 mol H₂SO₄) × (32.1g / 1 mol S) = 8.34 g S

∴7.48 g S is produced.

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