5 Steps to a 5 AP Chemistry

In another reaction, 40.0 g of Cl 2 and excess H 2 are combined. HCl will be produced.

How many grams of HCl will form?

H 2 (g) + Cl 2 (g) → 2 HCl(g)

Answer:

Limiting Reactants

In the examples above, one reactant was present in excess. One reactant was completely

consumed, and some of the other reactant would be left over. The reactant that is used up

first is called the limiting reactant(L.R.). This reactant really determines the amount of

product being formed. How is the limiting reactant determined? You can’t assume it is the

reactant in the smallest amount, since the reaction stoichiometry must be considered. There

are generally two ways to determine which reactant is the limiting reactant:

1. Each reactant, in turn, is assumed to be the limiting reactant, and the amount of prod-
   uct that would be formed is calculated. The reactant that yields the smallestamount of
   product is the limiting reactant. The advantage of this method is that you get to prac-
   tice your calculation skills; the disadvantage is that you have to do more calculations.


2. The moles of reactant per coefficient of that reactant in the balanced chemical equation
   is calculated. The reactant that has the smallest mole-to-coefficient ratio is the limiting
   reactant. This is the method that many use.

Let us consider the Haber reaction once more. Suppose that 50.0 g of nitrogen and 40.0 g of

hydrogen were allowed to react. Calculate the number of grams of ammonia that could be formed.

First, write the balanced chemical equation:

N 2 (g) +3 H 2 (g) →2 NH 3 (g)

Next, convert the grams of each reactant to moles:

Divide each by the coefficient in the balanced chemical equation. The smaller is the

limiting reactant:

For N 2 : 1.7848 mol N 2 /1 =1.7848 mol/coefficient limiting reactant

For H 2 : 19.8432 mol H 2 /3 =6.6144 mol/coefficient

Finally, base the stoichiometry of the reaction on the limiting reactant:

Anytime the quantities of more than one reactant are given it is probably a L.R. problem.

Let’s consider another case. To carry out the following reaction: P 2 O 5 (s)+3H 2 O(l) →

2H 3 PO 4 (aq) 125 g of P 2 O 5 and 50.0 g of H 2 O were supplied. How many grams of H 3 PO 4

may be produced?

50 0 1 2

1

2

2

.gN

1

mol N

28.014 g N

mol NH

m

××^23

ool N

g

1mol NH

gNH

3

3

2

17 0307

×=60 8

.

.

50 0

1

1

(^2) 1 7848

.

.

gN molN

28.014 g N

(^2) mol N
2

×= 22

2

2

gH molH

2.0158 g H

mol

40 0

1

1

(^2) 19 8432

.

=. H 2

40 0

(^12)
.gCl 2
mol Cl
70.906 g Cl
2 mol HC
2

()

⎛

⎝

⎜

⎞

⎠

⎟

ll

1molCl

gHCl

2 1molHCl

⎛

⎝

⎜

⎞

⎠

⎟

⎛

⎝

⎜

⎞

⎠

⎟=

36 461

41

.

.. 1 g H C l

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