SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 6

Stoichiometry (Chemical Calculations) and the Mole Concept

PERCENT YIELD OF A PRODUCT

In most stoichiometric problems, we assume that the results are exactly what we would theoretically expect. In reality, the resulting theoretical yield is rarely the actual yield. Why the actual yield of a reaction may be less than the theoretical yield occurs for many reasons. Some of the product is often lost during the purification or collection process.

Chemists are usually interested in the efficiency of a reaction. The efficiency is expressed by comparing the actual and the theoretical yields.

The percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%.

EXAMPLE:    Quicklime, CaO, is to be prepared by roasting 2.00 × 103 g of

limestone, CaCO3, according to the following reaction.

However, the actual yield of CaO is 1.05 × 103 g. What is the percent of yield?

The reaction is:

CaCO3(s) → CaO(s) + CO2(g)

The percent yield is found by using the above formula and inserting the actual and theoretical yields.

To find the theoretical yield, you solve the following:

Answers and Explanations to Embedded Chapter Practice Problems

After each type of problem in this chapter there are several problems to solve; therefore the chapter review does not include problems. The answer explanations for the problems previously presented are given below.

                  1. 

   Calculate the equation mass and volume so that the units match the ones above the equation.

Then

Using the mole method:

The mole relationship in the equation is:

2 moles H2O → 2H2 + 1 mole O2

Because 1 mol H2O = 18 g /mol

                                            

                  2.    

      Calculate the equation mass and volume so that the units match the ones above the equation.

Using the mole method:

The balanced equation shows that 4 mol Al needed 3 mol O2.

Then

Because there is 27 g in 1 mol Al,

       

                  3.    The equation:

Then

Using the mole method:

The balanced equation shows that 1 mol MnOneeded 1 mol Cl2.

Then

Because there is (54.9 + 32.0) or 86.9 g in 1 mol MnO2,

                  4.    The equation:

Comparing the amounts needed below the equation, you see that you need more Mg than O2. Therefore, the amount of Mg will be the limiting factor because it will be used up first. Because the equation masses indicate you must have more Mg than O2, there is an excess of oxygen.

Using the mole method:

The balanced equation shows that 2 mol Mg forms 2 mol MgO.

Then

Because there is (24 + 16) or 40. g of MgO in 1 mol MgO,

                  5.    The equation:

Set up the proportion using the coefficients of the substances that have something indicated above them as denominators. Gay-Lussac’s Law says that with reacting gases you can use the coefficients of the balanced equation to solve the problem.

                  6.    The equation:

Set up the proportion using the coefficients of the substances that have something indicated above them as denominators. Here again, because the reaction is of gases, you can use the coefficients of the balance equation to set up the proportion.