5 Steps to a 5 AP Chemistry

Answer:

1. Convert to moles:


2. Find the limiting reactant

The 1 mol and the 3 mol come from the balanced chemical equation. The 0.880 is smaller,

so this is the L.R.

3. Finish using the number of moles of the L.R.

Percent Yield

In the preceding problems, the amount of product calculated based on the limiting-

reactant concept is the maximum amount of product that could be formed from the given

amount of reactants. This maximum amount of product formed is called the theoretical

yield. However, rarely is the amount that is actually formed (the actual yield) the same as

the theoretical yield. Normally it is less. There are many reasons for this, but the principal

reason is that most reactions do not go to completion; they establish an equilibrium system

(see the Equilibrium chapter for a discussion of chemical equilibrium). For whatever

reason, not as much as expected is formed. The efficiency of the reaction can be judged by

calculating the percent yield. The percent yield(% yield) is the actual yield divided by the

theoretical yield, and the result is multiplied by 100% to generate percentage:

Consider the problem in which it was calculated that 60.8 g NH 3 could be formed. Suppose

that reaction was carried out, and only 52.3 g NH 3 was formed. What is the percent yield?

Let’s consider another percent yield problem in which a 25.0-g sample of calcium oxide is

heated with excess hydrogen chloride to produce water and 37.5 g of calcium chloride.

What is the percent yield of calcium chloride?

CaO(s) +2HCl(g) →CaCl 2 (aq) + H 2 O(l)

Answer:

(25.0 g CaO)

1molCaO

56.077 g CaO

⎛ 1molCa

⎝

⎜

⎞

⎠

⎟

CCl

1molCaO

110.984 g CaCl

1 mol CaCl

2 2

2

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜⎜

⎞

⎠

⎟=49.478 g CaCl^2

%%yield .%

52.3 g

60.8 g

=×= 100 86 0

%%yield

actual yield

theoretical yield

=× 100

0 880

298

.molPO

mol H PO

(^25) 1molPO
34
25

()

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

.. 0

172

4

gHPO

1molHPO

(^34) g
3

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=

0 880

0 880

278

092

25

.

.

.

.

mol

1mol

PO

mol

3mol

=

= 77 HO 2

125

1

142

(^5) 0 880
5
gPO.
mol P O
gP O
25 2 mol
2

()

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= PO

(50.0 g H O)

mol H O

18.0 g H O

m

25

2

2

2

1

278.

⎛

⎝

⎜

⎞

⎠

⎟= ool H O 2

Stoichiometry  93