Calculation of Amount of 50% H2SO4 Required to Decompose 25g of CaCO3
1. Chemical Reaction:
The chemical reaction between sulfuric acid (H2SO4) and calcium carbonate (CaCO3) can be represented as:
H2SO4 + CaCO3 → CaSO4 + CO2 + H2O
2. Molar Mass Calculation:
- Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol
- Molar mass of H2SO4 = 2(1.01 g/mol) (H) + 32.07 g/mol (S) + 4(16.00 g/mol) (O) = 98.08 g/mol
3. Stoichiometry:
From the balanced chemical equation, we see that 1 mole of CaCO3 reacts with 1 mole of H2SO4.
Therefore, 100.09 g of CaCO3 reacts with 98.08 g of H2SO4.
4. Calculation:
- Given mass of CaCO3 = 25g
- Using the molar mass of CaCO3, we can calculate the number of moles of CaCO3 present:
Moles of CaCO3 = Mass / Molar mass = 25g / 100.09 g/mol ≈ 0.25 mol
- Since the molar ratio of CaCO3 to H2SO4 is 1:1, 0.25 moles of H2SO4 are required.
- To find the mass of 50% H2SO4 needed, we use the molar mass of H2SO4:
Mass of H2SO4 = Moles of H2SO4 x Molar mass = 0.25 mol x 98.08 g/mol = 24.52 g
5. Conclusion:
Therefore, approximately 24.52 grams of 50% H2SO4 solution is required to decompose 25 grams of CaCO3. It is important to note that the actual amount needed may vary slightly due to factors such as purity of reagents and reaction conditions.