A balanced chemical equation contains a lot of information relating the individual chemical
entities, amounts of chemical entities, and the masses of reactants and products. A BALANCED
chemical equation is essential for all calculations involving reactants and products. When the
equation is balanced, we can derive stoichiometric relationships between reactants, products,
and reactant and products known as mole: mole ratios.
Balance the following equation:
C3H8 +
O2
1C3H8 + 5O2
 CO2

3CO2
+
H2O
+ 4H2O
From the balanced equation, the following stoichiometric relationships can be derived:
1 mole C 3 H 8
1 mole C 3 H 8
1 mole C 3 H 8 3 mole CO 2
5 mole O 2
4 mole H 2 O
;
;
;
;
;
5 mole O 2
1 mole C 3 H 8 3 mole CO 2 1 mole C 3 H 8 4 mole H 2 O
1 mole C 3 H 8
5 mole O 2 3 mole CO 2 5 mole O 2
4 mole H 2 O 3 mole CO 2 4 mole H 2 O
;
;
;
;
;
3 mole CO 2 5 mole O 2 4 mole H 2 O 5 mole O 2 4 mole H 2 O 3 mole CO 2
The mole : mole ratios can be used as conversion factors to calculate values for any species in
the chemical reaction.
Concept Test
If we use (consume) 15 moles of C3H8 how many moles of oxygen will be used?
This problem could not have been solved without first balancing the chemical equation. The
general approach for solving any problem that involved the chemical equation is:
1. Balance the chemical equation
2. Convert the given mass or entities of the first substance into moles
3. Use the correct mole : mole ratio from the balanced equation to calculate the moles of the
other substance
4. Convert the moles of the substance into mass or entities
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General Template for Chemical Reaction Conversions
given mass in grams of A
1 mol A
Y mol B molar mass in grams of B
 mass in grams of B
molar mass in grams of A X mol A
1 mol B
Concept Test
Solid copper (I) sulfide reacts in the presence of oxygen gas to form solid copper (I) oxide and
sulfur dioxide gas.
1. write a balanced chemical reaction with states of matter for each species
2. how many moles of oxygen gas will react with 10.0 moles of copper (I) sulfide?
3. how many grams of sulfur dioxide are formed when 10.0 moles of copper (I) sulfide react
with oxygen?
4. how many kilograms of oxygen are required to form 2.86 kg of copper (I) oxide?
Sometimes chemical reactions occur and we run out of one of the reagents. Thus, the formation
of the product is limited. Imagine if you wanted to make 4 batches of chocolate chip cookies,
and each batch requires 1 bag of chocolate chips. If you only have 3 bags of chocolate chips, you
cannot possible hope to make 4 batches of cookies! Thus, our product (the cookies) would be
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limited to 3 batches as that is the number of bag of chips that we have. It is the same with
chemical reactions. Once a reagent had been used up, no more product(s) can be made.
In choosing the limiting reactant, choose the reactant that will make (yield) the LOWEST
amount of product. The limiting reagent is NOT the reactant present in the fewest amount of
moles, it is the reactant that yields (forms) the fewest amount of moles of product. The limiting
reactant is NOT the reactant with the lowest mass. it is the reactant that forms the lower mass
of product.
Concept Test:
Liquids hydrazine (N2H4) and dinitrogen tetraoxide (N2O4) ignite to form nitrogen gas and
water vapor. How many grams of nitrogen gas will form when 1.00 x 102 grams of N2H4 and
2.00 x 102 grams of N2O4 are mixed? Which reactant is the limiting reagent? Is the limiting
reagent the same if we calculated the amount of water vapor formed instead? Prove your
answer mathematically.
2N2H4 (l) + 1N2O4 (l)  3N2 (g) + 4H2O (g)
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When you perform chemical reactions, ideally you will end up with a yield of 100%. It just does
not happen. Along they way, as chemical reactions proceed, product is lost. The theoretical
yield is the amount of product expected. You will CALCULATE the theoretical yield. What we
have been doing, calculating the amount of product given a certain number of moles of reactant
or grams of reactant IS calculating the theoretical yield. However, the actual yield is
determined by experimentation, and thus the value will be given to you.
% yield =
actual yield
x 100
theoretica l yield
Concept Test:
In the above problem you calculated the theoretical amount of N2 gas. What is the % yield if
you actually obtained 122 grams N2 instead?
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