Chemistry - Stoichiometry Notes



Chemistry - Stoichiometry Notes

I. Stoichiometry

A. Stoichiometry is the study of quantitative relationships that exist in chemical reactions.

B. Stoichiometry is very important in industry. When a specific amount of a product is needed stoichiometry can be used to determine exactly how much of the reactants are needed so that there is no waste.

C. Steps for solving stoichiometry problems.

1. Write a balanced equation.

2. If given grams in a problem convert them to moles.

3. Determine the ratio between the coefficient in front of the known or given substance and the coefficient in front of the unknown substance or what you are trying to find.

4. Multiply the moles of the known substance by the ratio (2nd no. goes on top).

5. The product of step 4 is the number of moles of the unknown substance. If you need grams, convert moles to grams.

D. Example - 6.0 grams of hydrogen gas reacts with excess oxygen gas to produce how many moles of water?

2 H2 + O2 ------------ 2 H2O

6.0 g. X 1 mole/2.0 g. = 3.0 moles

Ratio is 2 : 2 which can be reduced to 1 : 1

1 / 1 X 3.0 moles = 3.0 moles of water

E. Example - How many grams of zinc chloride could be produced from 19.5 grams of hydrochloric acid reacting with excess zinc?

Zn + 2 HCl ----------- ZnCl2 + H2

19.5 g. X 1 mole / 36.5 g. = .534 mole

Ratio is 2 : 1 so 1 / 2 X .534 mole = .267 mole

.267 mole X 136.4 g. / 1 mole = 36.4 grams

* Read in Prentice Hall Chemistry pages 353 thru 362.

II. Limiting Reactants

A. The limiting reactant is the reactant in a reaction that limits the amount of product that is formed.

B. When given the mass of two or more reactants, we need to first determine which reactant is in short supply in order to determine the amount of product that is formed.

C. To solve limiting reactant problems, follow the same steps as in stoichiometry problems except in step 2. In step 2 you determine the moles of both reactants and then use the moles of both to finish the problem.

D. Example - How many grams of iron II sulfide will be produced when 4.68 g. of iron reacts with 2.88 g. of sulfur?

Fe + S ------------- FeS

4.68 g. X 1 mole / 55.9 g. = .0837 mole

Ratio is 1:1 so .0837mole x 1/1 = .0837 mole

.0837 mole x 87.9 g = 7.36 g.

1 mole

------------------------------------------------------------

2.88 g. X 1 mole / 32.1 g. = .0897 mole

Ratio is 1:1 so .0897 x 1/1 = .0897 mole

.0837 mole X 87.9 g. = 7.88 g.

1 mole

We will use the smaller mass which is 7.36 g.

E. Example - When 1.3 grams of hydrogen gas reacts with 16 g. of oxygen gas what mass of water will be produced?

2 H2 + O2 --------- 2 H2O

1.3 g. X 1 mole / 2.0 g. = .65 mole

.65 moles x 18.0 g. = 12 g.

mole

---------------------------------------------------------------

16 g. X 1 mole / 32.0 g. = .50

Ratio is 1 : 2 or so 1 / 2 X .50 mole = 1.0 mole

1. moles x 18.0 g. = 18 g.

1 mole

We will use the smaller mass which is 12 g.

* Read in Prentice Hall Chemistry pages 368 thru 371.

III. Percent Yield

A. The percent yield is the amount of product actually formed as compared to the amount of product theoretically possible (calculated).

B. Used many times to evaluate your lab skills.

C. Percent Yield = Actual Yield / Calculated Yield X 100

Actual Yield - The amount actually produced when you carry out the reaction.

Calculated Yield - The amount of product calculated in a stoichiometry problem.

D. Example - Determine the percent yield for the reaction between 6.92 g. of potassium and 4.28 g. of oxygen if 7.36 g. of potassium oxide is produced?

4 K + O2 -------------- 2K2O

6.92 g. X 1 mole / 39.1 g. = .177 mole

4.28 g. X 1 mole / 32.0 g. = .134 mole K is limiting reactant because you need four moles for every one of oxygen.

Ratio is 4 : 2 or 2 : 1 1 / 2 X .177 mole = .0885 mole

.0885 mole X 94.2 g. / 1 mole = 8.34 g.

Percent Yield = 7.36 g. / 8.34 g. X 100 = 88.2 %

* Read in Prentice Hall Chemistry pages 372 and 375.

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