LIMITING QUANTITIES

10.3. LIMITING QUANTITIES

In Secs. 10.1 and 10.2, there was always sufficient (or excess) of all reactants except the one whose quantity was given. The quantity of only one reactant or product was stated in the problem. In this section, the quantities of more than one reactant will be stated. This type of problem is called a limiting-quantities problem.

How much Al 2 O 3 can be prepared from 2.0 mol of O 2 and 0.0 mol of Al? The first step, as usual, is to write the balanced chemical equation:

4 Al + 3 O 2 −→ 2 Al 2 O 3

It should be obvious that with no Al, there can be no Al 2 O 3 produced by this reaction. (This problem is not one which is likely to appear on examinations.) How much sulfur dioxide is produced by the reaction of 1.00 g S and all the oxygen in the atmosphere of the earth? (If you strike a match outside, do you really have to worry about not having enough oxygen to burn all the sulfur in the match head?) This problem has the quantity of each of two reactants stated, but it is obvious that the sulfur will be used up before the oxygen. It is also obvious that not all the oxygen will react! (Otherwise, we are all in trouble.) The problem is solved just as the problems in Sec. 10.2.

To solve a limiting-quantities problem in which the reactant in excess is not obvious, do as follows: If the reagents appear in a 1 mol: 1 mol ratio in the balanced chemical equation, you can tell immediately that the reagent present in lower number of moles is the one in limiting quantity. If they are not in a 1 mol: 1 mol ratio, an easy way to determine which is in limiting quantity is to divide the number of moles of each by the corresponding coefficient in the balanced chemical equation. The reagent with the lower quotient is in limiting quantity. Do not use these quotients for any further calculations. (In fact, it is useful to draw a line through them as soon as you see which reactant is in limiting quantity.) We can use the reactant in limiting quantity to determine the number of moles of each product that will be produced and the number of moles of the other reactant(s) that will be used up in the reaction. These steps are illustrated in Example 10.8 below.

In a limiting-quantities problem, you might be asked the number of moles of every substance remaining after the reaction. A useful way to calculate all the quantities is to use a table to do the calculations.

Step 1: Use the balanced equation as the heads of the quantities in the table, and label the rows “Present initially,” “Change due to reaction,” and “Present finally.” The initial quantities of reactants are entered in the first row. Zero is entered under the heading for any substances not present.

Step 2: Start the second row by entering the moles of limiting quantity—the same as the number of moles of that substance in the first row, since all of the limiting quantity is used up. The other entries in the second row are calculated using the technique of Sec. 10.1. The entries in row 2 will always be in the same mole ratio as the

coefficients in the balanced chemical equation. Step 3: Calculate the third row entries by adding the numbers of moles of products in the first two rows and

subtracting the numbers of moles of each reactant in the second row from the corresponding number in the first row. (If you ever get a negative number of moles in row 3, you have made a mistake somewhere.)

EXAMPLE 10.8. Calculate the number of moles of each substance present after 4.60 mol of aluminum is treated with 4.20 mol of oxygen gas and allowed to react.

Ans. The balanced equation and the left table entries are written first, along with the initial quantities from the statement of the problem:

Present initially:

Change due to reaction: Present finally:

Dividing 4.60 mol of Al by its coefficient 4 yields 1.15 mol Al; dividing 4.20 mol O 2 by 3 yields 1.40 mol O 2 . Since 1.15 is lower than 1.40, the Al is in limiting quantity. Therefore 4.60 mol is entered in the second row under Al. Note that the number of moles of Al present is greater than the number of moles of O 2 , but the Al is still in limiting quantity. The quantity of O 2 that reacts and of Al 2 O 3 that is formed is calculated as in Sec. 10.1:

2 = O 3.45 mol O

4.60 mol Al

2 4.60 mol Al

2 3 = 2.30 mol Al

4 mol Al

4 mol Al

These quantities are entered in line 2 also.

Present initially:

Change due to reaction:

Present finally:

The final line is calculated by subtraction for the reactants and addition for the product.

Present initially:

Change due to reaction:

Present finally:

0.00 mol

0.75 mol

2.30 mol EXAMPLE 10.9. How many moles of PbI 2 can be prepared by the reaction of 0.128 mol of Pb(NO 3 ) 2 and 0.206 mol NaI?

Ans. The balanced equation is

Pb(NO 3 ) 2 + 2 NaI −→ PbI 2 + 2 NaNO 3

The limiting quantity is determined:

0.128 mol Pb(NO 3 ) 2 0.206 mol NaI

1 2 = 2 0.103 mol NaI NaI is in limiting quantity.

= 0.128 mol Pb(NO 3 )

0.206 mol NaI

2 = 0.103 mol PbI

2 mol NaI

Note especially that the number of moles of NaI exceeds the number of moles of Pb(NO 3 ) 2 present, but the NaI is still in limiting quantity.

EXAMPLE 10.10. How many grams of Ca(ClO 4 ) 2 can be prepared by treatment of 12.0 g CaO with 102 g HClO 4 ? How many grams of excess reactant remains after the reaction?

Ans. This problem gives the quantities of the two reactants in grams; we must first change them to moles:

12.0 g CaO

= 0.214 mol CaO

56.0 g CaO

102.0 g HClO

1.02 mol HClO

100 g HClO

Now the problem can be done as in Example 10.8. All quantities are in moles. The balanced equation is

CaO + 2 HClO 4 −→ Ca(ClO 4 ) 2 + H 2 O

Present initially:

Change due to reaction:

Present finally:

0.214 mol Ca(ClO

51.1 g Ca(ClO )

4 2 produced

1 mol Ca(ClO

0.59 mol HClO 4 4 =

59 g HClO

1 mol HClO

If the quantities of both reactants are in exactly the correct ratio for the balanced chemical equation, then either reactant may be used to calculate the quantity of product produced. (If on a quiz or examination it is obvious that they are in the correct ratio, you should state that they are, so that your instructor will understand that you recognize the problem to be a limiting-quantities type problem.)

EXAMPLE 10.11. How many moles of lithium nitride, Li 3 N, can be prepared by the reaction of 0.600 mol Li and 0.100 mol N 2 ?

Since the ratio of moles of lithium present to moles of nitrogen present is 6 : 1, just as is required for the balanced equation, either reactant may be used.

3 N 2 = 0.200 mol Li 3 6 mol Li N 1 mol N

3 N = 0.200 mol Li

0.600 mol Li

or

0.100 mol N

(The first sentence of this answer should be stated on an examination.)