<h2>Steps to balance a chemical equation through inspection</h2>
<p>When balancing a chemical equation, there are a number of steps that need to be followed.</p>
<ol>
<li>
<p>Identify the reactants and the products in the reaction and write their chemical formulae.</p>
</li>
<li>
<p>Write the equation by putting the reactants on the left of the arrow and the products on the right.</p>
</li>
<li>
<p>Count the number of atoms of each element in the reactants and the number of atoms of each element in the products.</p>
</li>
<li>
<p>If the equation is not balanced, <a data-lesson="chemistry-104-summary-1" data-subject="chemistry" data-tutorial="physical-and-chemical-change" href="https://app.ulearngo.com/chemistry/physical-and-chemical-change/chemistry-104-summary-1">change</a> the coefficients of the molecules until the number of atoms of each element on either <a data-lesson="dissociation-in-water" data-subject="chemistry" data-tutorial="reactions-in-aqueous-solution" href="https://app.ulearngo.com/chemistry/reactions-in-aqueous-solution/dissociation-in-water">side</a> of the equation balance.</p>
</li>
<li>
<p>Check that the atoms are in fact balanced.</p>
</li>
<li>
<p>(we will look at this a little later): Add any extra details to the equation e.g. phase symbols.</p>
</li>
</ol>
<div class="ns-school-emp">
<h2>Example: Balancing Chemical Equations</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>Balance the following equation:</p>

\[\text{Mg} + \text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_{2}\]</div>
<div>
<h3>Step 1: Identify the reactants and products</h3>
<p>This has been done in the question.</p>
</div>
<div>
<h3>Step 2: Write the equation for the reaction</h3>
<p>This has been done in the question.</p>
</div>
<div>
<h3>Step 3: Count the number of atoms of each element in the reactants and products</h3>
<p>Reactants: \(\text{Mg} = 1 \text{ atom}\), \(\text{H} = 1 \text{ atom}\), \(\text{Cl} = 1 \text{ atom}\)</p>
<p>Products: \(\text{Mg} = 1 \text{ atom}\), \(\text{H} = 2 \text{ atoms}\), \(\text{Cl} = 2 \text{ atoms}\)</p>
</div>
<div>
<h3>Step 4: Balance the equation</h3>
<p>The equation is not balanced since there are two chlorine atoms in the product and only one in the reactants. If we add a coefficient of two to the \(\text{HCl}\) to increase the number of \(\text{H}\) and \(\text{Cl}\) atoms in the reactants, the equation will look like this:</p>

\[\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_{2}\]</div>
<div>
<h3>Step 5: Check that the atoms are balanced</h3>
<p>If we count the atoms on each side of the equation, we find the following:</p>
<p>Reactants: \(\text{Mg} = 1 \text{ atom}\), \(\text{H} = 2 \text{ atoms}\), \(\text{Cl} = 2 \text{ atoms}\)</p>
<p>Products: \(\text{Mg} = 1 \text{ atom}\), \(\text{H} = 2 \text{ atoms}\), \(\text{Cl} = 2 \text{ atoms}\)</p>
<p>The equation is balanced. The final equation is:</p>

\[\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_{2}\]</div>
</div>
</div>
<div class="ns-school-emp">
<h2>Example: Balancing Chemical Equations</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>Balance the following equation:</p>

\[\text{CH}_{4} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O}\]</div>
<div>
<h3>Step 1: Count the number of atoms of each element in the reactants and products</h3>
<p>Reactants: \(\text{C} = 1\), \(\text{H} = 4\), \(\text{O} = 2\)</p>
<p>Products: \(\text{C} = 1\), \(\text{H} = 2\), \(\text{O} = 3\)</p>
</div>
<div>
<h3>Step 2: Balance the equation</h3>
<p>If we add a coefficient of \(\text{2}\) to \(\text{H}_{2}\text{O}\), then the number of hydrogen atoms in the products will be \(\text{4}\), which is the same as for the reactants. The equation will be:</p>

\[\text{CH}_{4} + \text{O}_{2} \rightarrow \text{CO}_{2} + 2\text{H}_{2}\text{O}\]</div>
<div>
<h3>Step 3: Check that the atoms balance</h3>
<p>Reactants: \(\text{C} = 1\), \(\text{H} = 4\), \(\text{O} = 2\)</p>
<p>Products: \(\text{C} = 1\), \(\text{H} = 4\), \(\text{O} = 4\)</p>
<p>You will see that, although the number of <em>hydrogen</em> atoms now balances, there are more oxygen atoms in the products. You now need to repeat the previous step. If we put a coefficient of \(\text{2}\) in front of \(\text{O}_{2}\), then we will increase the number of oxygen atoms in the reactants by \(\text{2}\). The new equation is:</p>

\[\text{CH}_{4} + 2\text{O}_{2} \rightarrow \text{CO}_{2} + 2\text{H}_{2}\text{O}\]

<p>When we check the number of atoms again, we find that the number of atoms of each element in the reactants is the same as the number in the products. The equation is now balanced.</p>
</div>
</div>
</div>
<div class="ns-school-emp">
<h2>Example: Balancing Chemical Equations</h2>
<div class="ns-school-defn">
<h3>Question</h3>
<div>
<p>In our bodies, sugar \((\text{C}_{6}\text{H}_{12}\text{O}_{6})\) reacts with the oxygen we breathe in to produce <a data-lesson="carbon" data-subject="chemistry" data-tutorial="metals-metalloids-nonmetals" href="https://app.ulearngo.com/chemistry/metals-metalloids-nonmetals/carbon">carbon</a> dioxide, water and energy. Write the balanced equation for this reaction.</p>
</div>
<div>
<h3>Step 1: Identify the reactants and products in the reaction.</h3>
<p>Reactants: sugar \((\text{C}_{6}\text{H}_{12}\text{O}_{6})\) and oxygen \((\text{O}_{2})\)</p>
<p>Products: carbon dioxide \((\text{CO}_{2})\) and water \((\text{H}_{2}\text{O})\)</p>
</div>
<div>
<h3>Step 2: Write the equation</h3>

\[\text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O}\]</div>
<div>
<h3>Step 3: Count the number of atoms of each element in the reactants and in the products</h3>
<p>Reactants: \(\text{C} = 6\), \(\text{H} = 12\), \(\text{O} = 8\)</p>
<p>Products: \(\text{C} = 1\), \(\text{H} = 2\), \(\text{O} = 3\)</p>
</div>
<div>
<h3>Step 4: Balance the equation</h3>
<p>It is easier to start with carbon as it only appears once on each side. If we add a \(\text{6}\) in front of \(\text{CO}_{2}\), the equation looks like this:</p>

\[\text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{O}_{2} \rightarrow 6\text{CO}_{2} + \text{H}_{2}\text{O}\]

<p>Reactants: \(\text{C} = 6\), \(\text{H} = 12\), \(\text{O} = 8\)</p>
<p>Products: \(\text{C} = 6\), \(\text{H} = 2\), \(\text{O} = 13\)</p>
</div>
<div>
<h3>Step 5: Change the coefficients again to try to balance the equation.</h3>
<p>Let us try to get the number of hydrogens the same this time.</p>

\[\text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{O}_{2} \rightarrow 6\text{CO}_{2} + 6\text{H}_{2}\text{O}\]

<p>Reactants: \(\text{C} = 6\), \(\text{H} = 12\), \(\text{O} = 8\)</p>
<p>Products: \(\text{C} = 6\), \(\text{H} = 12\), \(\text{O} = 18\)</p>
</div>
<div>
<h3>Step 6: Now we just need to balance the oxygen atoms.</h3>

\[\text{C}_{6}\text{H}_{12}\text{O}_{6} + 6\text{O}_{2} \rightarrow 6\text{CO}_{2} + 6\text{H}_{2}\text{O}\]

<p>Reactants: \(\text{C} = 6\), \(\text{H} = 12\), \(\text{O} = 18\)</p>
<p>Products: \(\text{C} = 6\), \(\text{H} = 12\), \(\text{O} = 18\)</p>
</div>
</div>
</div>
<div class="note visit">
<h3><a href="http://phet.colorado.edu/en/simulation/balancing-chemical-equations">Recommended PhET Simulation: Balancing Chemical Equations</a></h3>
</div>
<section></section>

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Balancing Chemical Equations Continued

Chemistry generally explores the substances of which matter is composed, the investigation of their properties and reactions, and the use of such reactions to form new substances. 

Chemistry

Discover the different types of changes that matter can undergo and learn about the laws and principles that govern those changes, including the conservation of mass and volume relationships in gases, as well as the correct way of representing and balancing chemical equations.


Balancing Chemical Equations Continued

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