5 Steps to a 5 AP Chemistry 2019

(^92) › STEP 4. Review the Knowledge You Need to Score High
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: ()

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40.0 g Cl =

1 mol  Cl

70.906 g Cl

2 mol  HCl

1 mol Cl

36.461 g HCl

1 mol HCl

2 2 41.1 g HCl

22

Limiting Reactants

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

consumed, and some of the other reactant was 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 smallest amount of
   product is the limiting reactant. The advantage of this method is that you get to practice
   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:

×=

×=

50.0 g N

1

1 mol N

28.014 g N

1.7848 mol N

40.0 g H

1

1 mol  H

2.0158 g H

19.8432 mol H

2 2

2

2

22

2

2

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:

× ××=

50.0 g N

1

1 mol  N

28.014 gN

2 mol NH

1 mol  N

17.0307 g

1 mol NH

(^22) 60.8 g NH
2
3
23
3
Any time the quantities of more than one reactant are given it is probably an 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?
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