The equations that represent chemical reactions must show that the reaction obeys the law of conservation of mass. This is demonstrated with balanced chemical equations.
In the video, you learned that balancing equations involves adjusting the coefficients in the equation until the number of atoms of each element is equal on both sides of the equation. The steps are shown in this table.
| Step 1 | Make a list of elements on both sides of the equation. |
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| Step 2 | Multiply the coefficient and subscript to determine the number of atoms of each element. |
| Step 3 | Continue adjusting the coefficients until the number of atoms of each element are the same on both sides of the equation. |
Let's Practice
Practice writing balanced equations by completing this activity. Write the balanced equation to represent the chemical reaction described on each tab, then check your answer.
If you need a periodic table, click below to open an interactive periodic table or to download a PDF.
\(\text{Na}_{(\text{s})} + \text{HCl}_{(\text{aq})} \rightarrow \text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
\(\color{red}{2}\text{Na}_{(\text{s})} + \color{red}{2}\text{HCl}_{(\text{aq})} \rightarrow \color{red}{2}\text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
If you need help arriving at this answer, click the Solution button.
Remember that there is often more than one way to balance an equation. One way is shown here.
| Step 1: Make a list of elements on both sides of the equation. |
\(\text{Na}_{(\text{s})} + \text{HCl}_{(\text{aq})} \rightarrow \text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
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| Step 2: Multiply the coefficient and subscript to determine the number of atoms of each element. |
\(\text{Na}_{(\text{s})} + \text{HCl}_{(\text{aq})} \rightarrow \text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
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| Step 3: Continue adjusting the coefficients until the number of atoms of each element are the same on both sides of the equation. |
To balance the \(\text{H}\), a coefficient of \(2\) must be applied on the reactant side. \(\text{Na}_{(\text{s})} + \color{red}{2}\text{HCl}_{(\text{aq})} \rightarrow \text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
The coefficient must be applied to all elements in the compound. Therefore, \(\text{Cl}\) is no longer balanced; a coefficient of \(2\) must be applied on the product side. \(\text{Na}_{(\text{s})} + 2\text{HCl}_{(\text{aq})} \rightarrow \color{red}{2}\text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
The coefficient must be applied to all elements in the compound. Therefore, \(\text{Na}\) is no longer balanced; a coefficient of \(2\) must be applied on the reactant side. \(\color{red}{2}\text{Na}_{(\text{s})} + 2\text{HCl}_{(\text{aq})} \rightarrow 2\text{NaCl}_{(\text{aq})} + \text{H}_{2(\text{g})}\)
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\(\text{CaO}_{(\text{s})} + \text{SO}_{2(\text{g})} \rightarrow \text{CaSO}_{3(\text{s})}\)
The equation is already balanced!
If you need help arriving at this answer, click the Solution button.
Remember that there is often more than one way to balance an equation. One way is shown here.
| Step 1: Make a list of elements on both sides of the equation. |
\(\text{CaO}_{(\text{s})} + \text{SO}_{2(\text{g})} \rightarrow \text{CaSO}_{3(\text{s})}\)
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| Step 2: Multiply the coefficient and subscript to determine the number of atoms of each element. |
\(\text{CaO}_{(\text{s})} + \text{SO}_{2(\text{g})} \rightarrow \text{CaSO}_{3(\text{s})}\)
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| Step 3: Continue adjusting the coefficients until the number of atoms of each element are the same on both sides of the equation. | No coefficients are needed since the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side of the equation. |
Aqueous solutions of potassium hydroxide and phosphoric acid are combined to form a solution of potassium phosphate and water.
\(\color{red}{3}\text{KOH}_{(\text{aq})} + \text{H}_{3}\text{PO}_{4\ (\text{aq})} \rightarrow \text{ K}_{3}{\text{PO}}_{4\ (\text{aq})} + \color{red}{3} \text{H}_{2}\text{O}_{(\text{l})}\)
If you need help arriving at this answer, click the Solution button.
Remember that there is often more than one way to balance an equation. One way is shown here.
As mentioned in the tutorial video, it can be helpful to write water as \(\text{HOH}\) to visualize the hydroxide ion as a unit when balancing.
| Step 1: Make a list of elements on both sides of the equation. |
\(\text{KOH}_{\text{aq}} + \text{H}_{3}\text{PO}_{4\ \text{aq}} \rightarrow \text{K}_{3}\text{PO}_{4\ \text{aq}} + \color{red}{\text{HOH}}_{(\text{l})}\)
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| Step 2: Multiply the coefficient and subscript to determine the number of atoms of each element. |
\(\text{KOH}_{(\text{aq})} + \text{H}_{3}\text{PO}_{4\ (\text{aq})} \rightarrow \text{K}_{3}\text{PO}_{4\ (\text{aq})} + \text{HOH}_{(\text{l})} \)
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| Step 3: Continue adjusting the coefficients until the number of atoms of each element are the same on both sides of the equation. |
To balance the \(\text{K}\), a coefficient of \(3\) must be applied on the reactant side. \(\color{red}{3}\text{KOH}_{(\text{aq})} + \text{H}_{3}\text{PO}_{4\ (\text{aq})} \rightarrow \text{ K}3\text{PO}_{4\ (\text{aq})}\ + \text{HOH}_{(\text{l})}\)
The coefficient must be applied to all elements in the compound. Therefore, \(\text{OH}\) is no longer balanced; a coefficient of \(3\) must be applied on the product side. This will also balance the \(\text{H}\). \(3\text{KOH}_{(\text{aq})} + \text{H}_{3}\text{PO}_{4\ (\text{aq})} \rightarrow \text{K}_{3}\text{PO}_{4\ (\text{aq})} + \color{red}{3}\text{HOH}_{(\text{l})}\)
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Solid gold (III) sulfide reacts with hydrogen gas to form solid gold and hydrogen sulfide gas.
\(\text{Au}_{2}\text{S}_{3(\text{s})} + \color{red}{3}\text{H}_{2(\text{g})} \rightarrow \color{red}{2}\text{Au}_{(\text{s})} + \color{red}{3}\text{H}_{2}\text{S}_{(\text{g})}\)
If you need help arriving at this answer, click the Solution button.
Remember that there is often more than one way to balance an equation. One way is shown here.
| Step 1: Make a list of elements on both sides of the equation. |
\(\text{Au}_{2}\text{S}_{3(\text{s})} + \text{H}_{2(\text{g})} \rightarrow \text{Au}_{(\text{s})} + \text{H}_{2}\text{S}_{(\text{g})}\)
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| Step 2: Multiply the coefficient and subscript to determine the number of atoms of each element. |
\(\text{Au}_{2}\text{S}_{3(\text{s})} + \text{H}_{2(\text{g})} \rightarrow \text{Au}_{(\text{s})} + \text{H}_{2}\text{S}_{(\text{g})}\)
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| Step 3: Continue adjusting the coefficients until the number of atoms of each element are the same on both sides of the equation. |
To balance the \(\text{Au}\), a coefficient of \(2\) must be applied on the product side. \(\text{Au}_{2}\text{S}_{3(\text{s})} + \text{H}_{2(\text{g})} \rightarrow \color{red}{2}\text{Au}_{(\text{s})} + \text{H}_{2}\text{S}_{(\text{g})}\)
To balance the \(\text{S}\), a coefficient of \(3\) must be applied on the product side \(\text{Au}_{2}\text{S}_{3(\text{s})} + \text{H}_{2(\text{g})} \rightarrow 2\text{Au}_{(\text{s})} + 3\text{H}_{2}\text{S}_{(\text{g})}\)
The coefficient must be applied to all elements in the compound. Therefore, \(\text{H}\) is no longer balanced; a coefficient of three must be applied on the reactant side. \(\text{Au}_{2}\text{S}_{3\ (\text{s})} + \color{red}{3}\text{H}_{2(\text{g})} \rightarrow 2\text{Au}_{(\text{s})} + 3\text{H}_{2}\text{S}_{(\text{g})}\)
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Aqueous solutions of calcium chloride and sodium phosphate react to yield the precipitate calcium phosphate in sodium chloride solution.
\(3\text{CaCl}_{2(\text{aq})} + 2\text{Na}_{3}\text{PO}_{4(\text{aq})} \rightarrow \text{Ca}_{3}{(\text{PO}_{4})}_{2(\text{s})} + 6\text{NaCl}_{(\text{aq})}\)
If you need help arriving at this answer, click the Solution button.
Remember that there is often more than one way to balance an equation. One way is shown here.
| Step 1: Make a list of elements on both sides of the equation. |
\(\text{CaCl}_{2(\text{aq})} + \text{Na}_{3}\text{PO}_{4(\text{aq})} \rightarrow \text{Ca}_{3}{(\text{PO}_{4})}_{2(\text{s})} + \text{NaCl}_{(\text{aq})}\)
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| Step 2: Multiply the coefficient and subscript to determine the number of atoms of each element. |
\(\text{CaCl}_{2(\text{aq})} + \text{Na}_{3}\text{PO}_{4(\text{aq})} \rightarrow \text{Ca}_{3}{(\text{PO}_{4})}_{2(\text{s})} + \text{NaCl}_{(\text{aq})}\)
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| Step 3: Continue adjusting the coefficients until the number of atoms of each element are the same on both sides of the equation. |
To balance the \(\text{Ca}\), a coefficient of \(3\) must be applied on the reactant side. \(\color{red}{3}\text{CaCl}_{2(\text{aq})} + \text{Na}_{3}\text{PO}_{4(\text{aq})} \rightarrow \text{Ca}_{3}{(\text{PO}_{4})}_{2(\text{s})} + \text{NaCl}_{(\text{aq})}\)
The coefficient must be applied to all elements in the compound. \(\text{Cl}\) is not balanced; a coefficient of \(6\) must be applied on the product side. \(3 \text{ CaCl}_{2(\text{aq})} + \text{Na}_{3}\text{PO}_{4(\text{aq})} \rightarrow \text{Ca}_{3}{(\text{PO}_{4})}_{2(\text{s})} + \color{red}{6}\text{NaCl}_{(\text{aq})}\)
The coefficient must be applied to all elements in the compound. \(\text{Na}\) is not balanced; a coefficient of \(2\) must be applied on the reactant side. This will also balance the \(\text{PO}_{4}\). \(3\text{CaCl}_{2(\text{aq})} + \color{red}{2}\text{Na}_{3}\text{PO}_{4(\text{aq})} \rightarrow \text{Ca}_{3}{(\text{PO}_{4})}_{2(\text{s})} + 6\text{NaCl}_{(\text{aq})}\)
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