﻿ ﻿Balancing Chemical Equations - Compounds and Stoichiometry - Training MCAT General Chemistry Review

## MCAT General Chemistry Review

### Chapter 4: Compounds and Stoichiometry

4.4 Balancing Chemical Equations

Because chemical equations express how much and what types of reactants must be used to obtain a given quantity of product, it is of utmost importance that the reaction be balanced so as to reflect the laws of conservation of mass and charge. The mass of the reactants consumed must equal the mass of products generated. More specifically, one must ensure that the number of atoms of each element on the reactant side equals the number of atoms of that element on the product side. Stoichiometric coefficients, which are the numbers placed in front of each compound, are used to indicate the relative number of moles of a given species involved in the reaction. For example, the balanced equation expressing the combustion of nonane is:

C9H20 (g) + 14 O2 (g) → 9 CO2 (g) + 10 H2O (l)

The coefficients indicate that one mole of C9H20 gas must be reacted with fourteen moles of O2 gas to produce nine moles of carbon dioxide and ten moles of water. In general, stoichiometric coefficients are given as whole numbers.

MCAT EXPERTISE

It is unlikely that you will come across a question that explicitly asks you to balance an equation. However, you will need to recognize unbalanced reactions and quickly add the necessary coefficients. To balance a reaction, look at the number of atoms of each element and the charge on both sides (especially for oxidation–reduction reactions).

The steps taken to balance a chemical reaction are necessary to ensure that calculations regarding the reaction are performed correctly. Let’s review the steps involved in balancing a chemical equation, using an example.

Example:

Balance the following reaction:

C4H10 (l) + O2 (g) → CO2 (g) + H2O (l)

Method One:

First, balance the carbons (4 on reactant side) in the products.

C4H10 (l) + O2 (g) → 4 CO2 (g) + H2O (l)

Then, balance the hydrogens (10 on reactant side) in the products.

C4H10 (l) + O2 (g) → 4 CO2 (g) + 5 H2O (l)

Next, balance the oxygens (now 13 on product side) in the reactants. Finally, produce a whole number ratio. In this case, double each coefficient.

2 C4H10 (l) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l)

Finally, check that all of the elements and the total charges are balanced correctly. If there is a difference in total charge between the reactants and products, then the charge will also have to be balanced. (Instructions for balancing charge in oxidation–reduction reactions are found in Chapter 11 of MCAT General Chemistry Review.)

Method Two:

First, if in doubt, take a guess. Assume there are 4 of the first reactant and balance the carbons appropriately.

4 C4H10 (l) + O2 (g) → 16 CO2 (g) + H2O (l)

Second, balance the hydrogens (40 on reactant side) in the products.

4 C4H10 (l) + O2 (g) → 16 CO2 (g) + 20 H2O (l)

Third, balance the oxygens (now 52 on product side) in the reactants.

4 C4H10 (l) + 26 O2 (g) → 16 CO2 (g) + 20 H2O (l)

Fourth, produce the simplest whole number ratio through the greatest common factor. In this case, divide each side by 2.

2 C4H10 (l) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l)

Finally, check that all of the elements and the total charges are balanced correctly. Notice that both methods produce a multiple of our final answer. These ratios are not incorrect, but in terms of the stoichiometry one performs on the MCAT, the simpler the numbers are, the easier calculations will become.

KEY CONCEPT

When balancing equations, focus on the least represented elements first and work your way to the most represented element of the reaction (usually oxygen or hydrogen). If you,re stuck, take a guess for the coefficient of the first reactant and balance the remainder appropriately.

MCAT Concept Check 4.4:

Before you move on, assess your understanding of the material with this question.

1. Balance the following reactions:

· Fe + Cl2 → FeCl3

· Zn + HCl → ZnCl2 + H2

· C5H12 + O2 → CO2 + H2O

· Pb(NO3)2 + AlCl3 → PbCl2 + Al(NO3)3

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