Question

Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodium carbonate must be added to neutralize 5.04Ã103 kg of sulfuric acid solution? express your answer with the appropriate units.

Answer 1

5.45 x 10^3 kg of sodium carbonate is needed to neutralize 5.04 kg of sulfuric acid. For this, I will assume you have pure H2SO4. So first, you need to calculate the molar mass of H2SO4 and Na2CO3. Lookup the atomic weights of all the elements involved. Atomic weight of Sodium = 22.989769 Atomic weight of Sulfur = 32.065 Atomic weight of Carbon = 12.0107 Atomic weight of Oxygen = 15.999 Atomic weight of Hydrogen = 1.00794 Molar mass of H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass of Na2CO3 = 2 * 22.989769 + 12.0107 + 3 * 15.999 = 105.987238 g/mol The balanced equation for the reaction of Na2CO3 with H2SO4 is Na2CO3 + H2SO4 => Na2SO4 + CO2 + H2O so for every mole of sulfuric acid to be neutralized, you need 1 mole of sodium carbonate. You can determine the number of moles of sulfuric acid you have and then calculate the mass of that many moles of sodium carbonate. But, there's an easier way. Just use the relative mass differences between sodium carbonate and sulfuric acid. So 105.987238 g/mol / 98.07688 g/mol = 1.080655 So that means for every kg of sulfuric acid, you need 1.080655 kg of sodium carbonate. Now do the multiplication. 5.04 x 10^3 kg * 1.080655 = 5.4465 x 10^3 kg. Since you only have 3 significant figures for your data, round the result to 3 significant figures, giving 5.45 x 10^3 kg