# 3.10 Calculations Involving a Limiting Reactant

```3.10 Calculations
Involving a Limiting
Reactant
Definitions
• Limiting-reactant principle – The maximum
amount of product possible from a reaction is
determined by the amount of reactant present
in the least amount, based on its reaction
coefficient and molecular weight.
• Limiting reactant – the reactant present in a
reaction in the least amount, based on its
reaction coefficients and molecular weight. It
is the reactant that determines the maximum
amount of product that can be formed.
3–2
Chemical Reaction for CO2
Reactants
Product
O2 (g)
CO2 (g)
C(s)
3–3
Mixture of reactants
+
C(s)
O2(g)
CO2(g) + unreacted O2
3–4
Learning Check
For the balanced equation shown below, what would be
the limiting reagent if 79.6 grams of NO were reacted
with 59.5 grams of O2?
2NO+O2=>2NO2
; NO or O2
3–5
Solution
To answer this question, calculate the grams of NO2
needed to react fully with 79.6 grams of NO and 59.5
grams of O2, by using the balanced equation.
79.6 g X 1 mol of NO X 2 mol of NO2 X 46 grams of NO2
of NO 30.0 g of NO 2 mol of NO
1 mol of NO2
= 122.05 grams of NO2
59.5 g X 1 mol of O2 X 2 mol of NO2 X 46 grams of NO2
of O2
32.0 g of O2 1 mol of O2
1 mol of NO2
= 171.06 grams of NO2
There is less NO2 with NO than O2, therefore,
the limiting reactant is NO
3–6
Learning Check
For the balanced equation shown below, if 19.1 grams of
CH5N were reacted with 88.1 grams of O2, how many
grams of CO2 would be produced, using the limited
reactant to determine the quantity of a product that
should be produced ?
4CH5N + 11O2 => 4CO2 + 10H2O + 4NO
3–7
Solution
To answer this question, calculate the grams of NO2
needed to react fully with 19.1 grams of CH5N and 88.1
grams of O2, by using the balanced equation.
19.1 g of X 1 mol of CH5N X 4 mol of CO2 X 44 g of CO2
CH5N
31.0 g of CH5N 4 mol of CH5O 1 mol of NO2
= 21.1 grams of CO2
88.1 g X 1 mol of O2 X 4 mol of CO2 X 44 grams of CO2
of O2
32.0 g of O2 11 mol of O2
1 mol of CO2
= 44.1 grams of CO2
There is 21.1 g of CO2 produced with CH5N
than O2, which is the limiting reactant
3–8
Percent Yield
• Percentage yield – the percentage of the
theoretical amount of a product actually
produced by a reaction.
• Actual yield – the mass product obtained in an
experiment.
• Theoretical yield – the mass calculated to give
the maximum amount of product.
% yield = Actual yield
X 100
Theoretical yield
3–9
Learning Check
A chemist wants to produce urea (N2CH4O) by reacting
ammonia (NH3) and carbon dioxide (CO2). The balanced
equation for the reaction is
2NH3(g) + CO2 (g) Æ N2CH4O(s) + H2O(l)
The chemist reacts 5.11 g NH3 with excess CO2 and
isolates 3.12 g of solid N2CH4O. Calculate the
percentage yield of the experiment.
3–10
Solution
To answer this question, calculate first the theoretical
yield of N2CH4O that should be made. Then use the
actual yield to calculate the percentage yield.
5.11 g of X 1 mol of NH3 1 mol of N2CH4O 60.1 g of N2CH4O
X
NH3 17.0 g of NH3 X 2 mol of NH3
1 mol of N2CH4O
= 9.03 grams of N2CH4O is the theoretical yield
The actual yield was 3.12 g of N2CH4O, so the % yield is
% yield = Actual yield
X 100 = 3.12 g X 100 = 34.6%
Theoretical yield
9.03 g
3–11
Learning Check
• Methanol (CH3OH), also called methyl
alcohol, is the simplest alcohol. It is used as
a fuel in race cars and is a potential
replacement for gasoline. Methanol can be
manufactured by combination of gaseous
carbon monoxide and hydrogen. Suppose
68.5 kg CO(g) is reacted with 8.60 kg H2(g).
Calculate the theoretical yield of methanol. If
3.57x104 g CH3OH is actually produced, what
is the percent yield of methanol?
3–12
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