5 Steps To Balancing The Equation For H2so4 And Naoh

To balance the equation for the reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH), you need to follow a systematic approach. Balancing chemical equations is an essential skill in chemistry, ensuring that the law of conservation of mass is satisfied. In this article, we will outline 5 detailed steps to effectively balance the equation for the neutralization reaction between H₂SO₄ and NaOH.

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Understanding the Reactants and Products

Before diving into the balancing process, it's crucial to understand the reactants involved in the reaction.

Reactants:

• Sulfuric Acid (H₂SO₄): A strong acid that completely dissociates in solution.
• Sodium Hydroxide (NaOH): A strong base that also dissociates fully in solution.

Products:

The expected products of this neutralization reaction are water (H₂O) and sodium sulfate (Na₂SO₄).

In a simple form, the reaction can be written as:

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

Now, let's proceed with the steps to balance this equation.

Step 1: Write the Unbalanced Equation

Start by writing the unbalanced equation based on the reactants and products identified.

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

Step 2: Count the Atoms of Each Element

Next, count the number of atoms for each element on both the reactants and products sides.

Reactants:

• H: 2 from H₂SO₄ and 1 from NaOH = 3 H
• S: 1 from H₂SO₄ = 1 S
• O: 4 from H₂SO₄ and 1 from NaOH = 5 O
• Na: 1 from NaOH = 1 Na

Products:

• H: 2 from H₂O = 2 H
• S: 2 from Na₂SO₄ = 1 S
• O: 4 from Na₂SO₄ and 1 from H₂O = 5 O
• Na: 2 from Na₂SO₄ = 2 Na

Now, let's summarize these counts in a table for clarity:

<table> <tr> <th>Element</th> <th>Reactants</th> <th>Products</th> </tr> <tr> <td>H</td> <td>3</td> <td>2</td> </tr> <tr> <td>S</td> <td>1</td> <td>1</td> </tr> <tr> <td>O</td> <td>5</td> <td>5</td> </tr> <tr> <td>Na</td> <td>1</td> <td>2</td> </tr> </table>

Step 3: Balance the Sodium (Na) Atoms

From our count, we see that sodium is unbalanced. There are 1 Na in the reactants and 2 Na in the products. To balance sodium, we need to place a coefficient of 2 before NaOH.

H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O

Now, let's recount the atoms:

Reactants after balancing:

• H: 2 from H₂SO₄ and 2 from 2 NaOH = 4 H
• S: 1 from H₂SO₄ = 1 S
• O: 4 from H₂SO₄ and 2 from 2 NaOH = 6 O
• Na: 2 from 2 NaOH = 2 Na

Products:

• H: 2 from H₂O = 2 H
• S: 2 from Na₂SO₄ = 1 S
• O: 4 from Na₂SO₄ and 1 from H₂O = 5 O
• Na: 2 from Na₂SO₄ = 2 Na

Now the table updates:

<table> <tr> <th>Element</th> <th>Reactants</th> <th>Products</th> </tr> <tr> <td>H</td> <td>4</td> <td>2</td> </tr> <tr> <td>S</td> <td>1</td> <td>1</td> </tr> <tr> <td>O</td> <td>6</td> <td>5</td> </tr> <tr> <td>Na</td> <td>2</td> <td>2</td> </tr> </table>

Step 4: Balance the Hydrogen (H) Atoms

With sodium balanced, we now focus on hydrogen. The reactants have 4 H, while the products only have 2 H. To balance this, we can place a coefficient of 2 before H₂O in the products.

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

Let's recount the atoms after this change:

Reactants:

• H: 4 from H₂SO₄ and 2 from 2 NaOH = 4 H
• S: 1 from H₂SO₄ = 1 S
• O: 4 from H₂SO₄ and 2 from 2 NaOH = 6 O
• Na: 2 from 2 NaOH = 2 Na

Products after balancing:

• H: 4 from 2 H₂O = 4 H
• S: 2 from Na₂SO₄ = 1 S
• O: 4 from Na₂SO₄ and 2 from 2 H₂O = 6 O
• Na: 2 from Na₂SO₄ = 2 Na

Updating the table once again gives us:

<table> <tr> <th>Element</th> <th>Reactants</th> <th>Products</th> </tr> <tr> <td>H</td> <td>4</td> <td>4</td> </tr> <tr> <td>S</td> <td>1</td> <td>1</td> </tr> <tr> <td>O</td> <td>6</td> <td>6</td> </tr> <tr> <td>Na</td> <td>2</td> <td>2</td> </tr> </table>

Step 5: Verify That All Atoms Are Balanced

Now that the equation reads:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

Let's conduct a final check of each element:

Final Counts

• H: 4 (Reactants) = 4 (Products)
• S: 1 (Reactants) = 1 (Products)
• O: 6 (Reactants) = 6 (Products)
• Na: 2 (Reactants) = 2 (Products)

Conclusion

All elements are balanced. Therefore, the balanced chemical equation for the reaction between sulfuric acid and sodium hydroxide is:

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

This balanced equation accurately represents the stoichiometry of the neutralization reaction. Understanding how to balance chemical equations is not only pivotal for theoretical chemistry but also plays an important role in practical applications such as titrations and industrial processes.

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