Stoichiometry And Balancing Reactions - Chemistry LibreTexts

Reactants to Products

A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation:

\[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber \]

In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium (\(Na\)), hydrogen (\(H\)), and chloride (\(Cl\)) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements.

Stoichiometric Coefficients

In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation:

\[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber \]

we can determine that 2 moles of \(HCl\) will react with 2 moles of \(Na_{(s)}\) to form 2 moles of \(NaCl_{(aq)}\) and 1 mole of \(H_{2(g)}\). If we know how many moles of \(Na\) reacted, we can use the ratio of 2 moles of \(NaCl\) to 2 moles of Na to determine how many moles of \(NaCl\) were produced or we can use the ratio of 1 mole of \(H_2\) to 2 moles of \(Na\) to convert to \(NaCl\). This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems.

Example 1

Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.

\[\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber \]

Solution

Start by counting the number of atoms of each element.

UNBALANCED

Element	Reactant (# of atoms)	Product (# of atoms)
Pb	1	1
O	8	9
H	6	2
S	1	2

The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of \(H_2SO_4\) to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of \(H_2O\) where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced.

\[\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber \]

BALANCED

Element	Reactant (# of atoms)	Product (# of atoms)
Pb	1	1
O	12	12
H	8	8
S	2	2

Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last.

A balanced equation ultimately has to satisfy two conditions.

1. The numbers of each element on the left and right side of the equation must be equal.
2. The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.