Determining the Required Grams of Sulfuric Acid to React with Sodium Carbonate

Understanding the stoichiometry involved in a chemical reaction can help us calculate the amount of reactants necessary for a complete reaction. In this article, we will explore how to determine the grams of sulfuric acid (H2SO4) needed to completely react with a sodium carbonate (Na2CO3) solution containing 10.6 grams of the salt.

Chemical Reaction and Balanced Equation

The chemical reaction involving sodium carbonate and sulfuric acid is as follows:

(Na2CO3 H2SO4 → Na2SO4 CO2 H2O)

This is the balanced chemical equation for the reaction between sodium carbonate and sulfuric acid. It shows a 1:1 molar ratio between the two reactants, meaning one mole of sodium carbonate reacts with one mole of sulfuric acid.

Step-by-Step Calculation

Step 1: Calculate Moles of Sodium Carbonate

Calculation of Molar Mass of Na2CO3: Na (Sodium): 45.98 g/mol (2 atoms)C (Carbon): 12.01 g/molO (Oxygen): 16.00 g/mol (3 atoms)

Total molar mass of Na2CO3 105.99 g/mol

To find the moles of Na2CO3 in 10.6 grams:

Moles of Na2CO3 10.6 g / 105.99 g/mol ≈ 0.100 mol

Step 2: Calculate Moles of Sulfuric Acid

Since the balanced equation shows a 1:1 molar ratio, the moles of sulfuric acid required will also be approximately 0.100 mol.

Step 3: Calculate Grams of Sulfuric Acid

Calculation of Molar Mass of H2SO4: H (Hydrogen): 1.01 g/mol (2 atoms)S (Sulfur): 32.07 g/molO (Oxygen): 16.00 g/mol (4 atoms)

Total molar mass of H2SO4 98.09 g/mol

Now, to find the mass of H2SO4 needed:

Mass of H2SO4 0.100 mol × 98.09 g/mol ≈ 9.81 g

Conclusion

To fully react with a 10.6 grams of sodium carbonate, approximately 9.81 grams of sulfuric acid are required. This derivation is based on the stoichiometry of the reaction and the molar masses of the reactants involved.

While the initial calculation suggests that 9.8 grams of sulfuric acid is needed, the more precise calculation based on molar ratios leads to 9.81 grams.