Balancing Chemical Equations (Simplified): Videos & Practice Problems

Balancing chemical equations is essential to ensure that the law of conservation of mass is upheld, meaning the type and number of atoms must be equal on both sides of the reaction arrow. In a balanced equation, the numbers preceding the chemical formulas, known as coefficients, indicate how many molecules of each substance are involved in the reaction. For instance, in a balanced equation, coefficients such as 2, 1, and 2 can be used to represent the quantities of reactants and products.

When balancing, it is crucial to distribute these coefficients correctly. For example, if a coefficient of 2 is applied to a compound containing hydrogen, it means that the total number of hydrogen atoms is doubled. If there are already 2 hydrogen atoms in the compound, the calculation would be 2 (coefficient) times 2 (existing atoms), resulting in 4 hydrogen atoms. Similarly, if the coefficient of 1 is applied to a compound with 2 oxygen atoms, the total remains 2, as 1 times 2 equals 2.

On the product side, if a coefficient of 2 is applied to a compound with 2 hydrogen atoms, the total becomes 4 hydrogen atoms (2 times 2). If there is also 1 oxygen atom in the product, the total for oxygen would be 2 (from the coefficient) times 1, resulting in 2 oxygen atoms. Thus, for a balanced equation, the types of atoms and their respective quantities must match on both sides. In this example, there are 4 hydrogen atoms and 2 oxygen atoms on each side, confirming that the equation is balanced.

Understanding how to balance chemical equations is fundamental in chemistry, as it allows for accurate representation of chemical reactions and ensures that mass is conserved throughout the process.

To balance the chemical equation for the combustion of butane (C₄H₁₀), we start by identifying the reactants and products involved. The reactants are butane and oxygen (O₂), while the products are water (H₂O) and carbon dioxide (CO₂).

Initially, we list the number of each type of atom present in the reactants and products. In butane, there are 4 carbon atoms, 10 hydrogen atoms, and 2 oxygen atoms. In the products, we have 1 carbon atom, 2 hydrogen atoms, and a total of 3 oxygen atoms (1 from H₂O and 2 from CO₂). This shows that the equation is unbalanced.

To balance the equation, we proceed by adjusting the coefficients. Starting with carbon, we place a coefficient of 4 in front of CO₂ to balance the carbon atoms, resulting in 4 carbon atoms on both sides. This adjustment also changes the oxygen count to 8 from CO₂ plus 1 from H₂O, totaling 9 oxygen atoms on the product side.

Next, we address the hydrogen atoms. To balance the 10 hydrogen atoms from butane, we place a coefficient of 5 in front of H₂O, which gives us 10 hydrogen atoms on the product side. This adjustment increases the total oxygen count to 5 from H₂O plus 8 from CO₂, resulting in 13 oxygen atoms on the product side.

Now, we need to balance the oxygen atoms. Since we have 2 oxygen atoms in each O₂ molecule, we look for a coefficient that will yield 13 total oxygen atoms. A coefficient of 6.5 in front of O₂ achieves this, as 6.5 times 2 equals 13. However, since coefficients must be whole numbers, we multiply the entire equation by 2 to eliminate the fraction.

After multiplying, the balanced equation becomes:

2 C₄H₁₀ + 13 O₂ → 10 H₂O + 8 CO₂

Thus, the coefficients for the balanced equation are 2, 13, 10, and 8, respectively, for butane, oxygen, water, and carbon dioxide.