Overview: In this tutorial, the fundamentals of balancing chemical reactions are reviewed.

Skills:

  • Balancing Chemical Equations

New terms:

  • Stoichiometry
  • Stoichiometric Coefficients

In chemistry it is very important to understand the relationship between reactants and products in a reaction. Stoichiometry is exactly that. It is the quantitative relation between the number of moles (and therefore mass) of various products and reactants in a chemical reaction.
Chemical reactions must be balanced, or in other words, must have the same number of various atoms in the products as in the reactants.

If a chemical reaction is not balanced, no information about the relationship between products and reactants can be derived. So the first thing to do when you see a chemical reaction is to balance it. We balance reactions by adding coefficients in front of the reactants and products. These coefficients are the stoichiometric coefficients.

General Guidelines for Balancing Simple Equations

  1. 1. Assign a “1” as the coefficient for the most complex species (the one whose chemical formula has the greatest number of different elements).
  2. 2. Balance any single-element species last.
  3. 3. Eliminate fractional coefficients (although this is not necessary).
  4. 4. Add coefficients only; do not change the chemical formulas.
  5. 5. There must be the same number of atoms on the left and right sides of the chemical reaction.

These are just guidelines, not rules. Therefore, sometimes it may be necessary to deviate from these general guidelines.

Example 1.

Balance the chemical reaction.

This equation is not balanced since there are more N and O atoms on the left side of the equation. Let’s start by using the guidelines. Assign a stoichiometric coefficient of 1 to the most complex compound, NO.

Now we can balance the remaining single-element compounds. In order to do this we will need to use fractional coefficients.

We can get rid of the fractional coefficients by multiplying by 2 even though this is a perfectly acceptable balanced chemical equation.

At the very beginning of this problem, perhaps you could see this was the answer. If you can see the balanced equation by sight, you don’t need to go by the guidelines. Remember they are only guidelines to help if you run into trouble. You can see by simply adding a 2 in front of NO, we violate the first guideline even though it leads us to a balanced equation.

Example 2.

Balance the given chemical reaction.

This one may not be as easy to see the final answer so we will use the guidelines to balance the equation. N2O3 is the most complex species so we will add a 1 for its coefficient.

Now we can balance the remaining single element species. In order to balance the number of atoms we need 2 atoms of N and 3 atoms of oxygen on the left side of the equation. There are already 2 atoms of N so we can add 3/2 in front of oxygen to get 3 atoms of oxygen.

The equation is now balanced. However, we can get rid of the fractional coefficient by again multiplying by 2.

Notice that in these two examples N2 and O2 react with a different stoichiometry to obtain different products. Is it necessary for the number of moles of the reactants to be equal to the number of moles of products? Answer

Not only does the stoichiometry tell us the mole relation between product and reactants but it will also tell us the mass relation.

Example 3.

How many grams of CO2 are produced by the complete combustion of 1.00 g of glucose (C6H12O6)? Remember that a combustion reaction is one in which the reactant (glucose in this case) reacts with O2 to produce CO2 and H2O. So first let’s write the skeleton equation (unbalanced equation).

Use the guidelines to balance the equation. We will assign a 1 to glucose.

We will balance the oxygen last because it is contained in a single-element species (O2). This means we need to balance the C and H atoms next. So we need 6 atoms of C and 12 atoms of H on the right side of the reaction.

So now we have 6 atoms of C and 6 atoms of H2 (which accounts for twelve H atoms). Now we can balance the oxygen. We have a total of 18 oxygen atoms on the right side. In order to have 18 atoms of oxygen on the left side we will need to assign a 6 to O2.

What does this really say? We need one mole of glucose to react with 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O. But the question asks about the mass of CO2 produced not the number of moles. We can use the molecular mass to convert.

The conversion factor of 6 moles of CO2 per mole of glucose comes from the balanced chemical equation (or the correct stoichiometry).


Summary

Now you should be able to balance chemical equations and understand what relations a balanced equation can reveal.


Practice Problems

  1. 1. Balance the following chemical reactions: Hint

    a. CO + O2 –>CO2

    b. KNO3 –> KNO2 + O2

    c. O3–> O2

    d. NH4NO3 –>N2O + H2O

    e. CH3NH2 + O2 –> CO2 + H2O + N2 Hint

    f. Cr(OH)3 + HClO4 –> Cr(ClO4)3 + H2O

  2. 2. Write the balanced chemical equations of each reaction:

    a. Calcium carbide (CaC2) reacts with water to form calcium hydroxide (Ca(OH)2) and acetylene gas (C2H2).

    b. When potassium chlorate (KClO3) is heated, it decomposes to form KCl and oxygen gas (O2).

    c. C6H6 combusts in air. Hint

    d. C5H12O combusts in air.

  3. 3. Given the following reaction: Na2S2O3 + AgBr –>NaBr + Na3[Ag(S2O3)2]

    a. How many moles of Na2S2O3 are needed to react completely with 42.7 g of AgBr?

    b. What is the mass of NaBr that will be produced from 42.7 g of AgBr?

    Hint

  4. 4. From the reaction: B2H6 + O2 –> HBO2 + H2O

    a. What mass of O2 will be needed to burn 36.1 g of B2H6?

    b. How many moles of water are produced from 19.2 g of B2H6?

  5. 5. Calculate the mass (in kg) of water produced from the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasoline is 0.79 g/mL.

  6. 6. One mole of aspartame (C14H18N2O5) reacts with two moles of water to produce one mole of aspartic acid (C4H7NO4), one mole of methanol (CH3OH) and one mole of phenylalanine.

    a. What is the molecular formula of phenylalanine? Hint

    b. What mass of phenylalanine is produced from 378 g of aspartame?

  7. 7. KO2 is used in a closed-system breathing apparatus. It removes carbon dioxide and water from exhaled air. The reaction for the removal of water is: KO2 + H2O –> O2 + KOH. The KOH produced is used to remove carbon dioxide by the following reaction: KOH + CO2 –> KHCO3.

    a. What mass of KO2 produces 235 g of O2?

    b. What mass of CO2 can be removed by 123 g of KO2Hint


Answer Key