## SAT Subject Test Chemistry

__PART 2__

__REVIEW OF MAJOR TOPICS__

__CHAPTER 6__

__Stoichiometry (Chemical Calculations) and the Mole Concept__

__MASS–MASS PROBLEMS__

A typical problem concerning just mass relationships is as follows:

**EXAMPLE 1:** What mass of oxygen can be obtained from heating 100.

grams of potassium chlorate?

1st step. Write the balanced equation for the reaction.

2KClO_{3} → 2KCl + 3O_{2}

2nd step. Write the given quantity and the unknown quantity above the appropriate substances.

3rd step. Calculate the equation mass for each substance that has something

indicated above it, and write the results under the substances. Note that the units above and below *must *match.

**TIP **

Put what is given above the equation.

Put the calculated amount below.

The **units must match!**

**USING THE PROPORTION METHOD:**

4th step. Form the proportion.

5th step. Solve for *x*.

*x *= 39.3 g of O_{2}

**USING THE FACTOR-LABEL METHOD:**

From the 3rd step on you would proceed as follows:

4th step.

The equation indicates that 245 g KClO_{3} yields 96 g of O_{2}. Therefore multiplying the given quantity by a factor made up of these two quantities arranged appropriately so that the units of the answer remain uncanceled will give the answer to the problem.

**TIP **

After you cancel all similar units above and below, you should have remaining the units for your answer.

**USING THE MOLE METHOD:**

To solve this problem you would proceed as follows from the 2nd step:

**EXAMPLE 2: **What mass of potassium hydroxide is needed to neutralize 20.0

grams of sulfuric acid?

**USING THE PROPORTION METHOD:**

**USING THE FACTOR-LABEL METHOD:**

**USING THE MOLE METHOD:**

**TRY THESE MASS–MASS PRACTICE PROBLEMS:**

__3.__ What mass of manganese dioxide is needed to react with an excess of hydrochloric acid so that 100. grams of chlorine is liberated?

Ans.: 122.5 g*

__4.__ A 20.0 gram sample of Mg is burned in 20.0 grams of O_{2}. How much MgO is formed? (Hint: Determine which is in excess.)

Ans. 33.3 g*