Question

Acid spills are often neutralized with sodium carbonate. For example

Na CO2 (s) + H2SO4 (aq) → Na2SO4 (aq) + CO2(g) + H20 (1)
An instructor dropped a 2.50 L bottle of 18.0 M H2SO4 on a cement floor.
How much sodium carbonate would be required to neutralize it?

Answer 1

4770 grams of sodium carbonate

Step-by-step explanation:

The reaction that takes place between sodium carbonate and sulfuric acid is:

• Na₂CO₃ (s) + H₂SO₄ (aq) → Na₂SO₄ (aq) + CO₂(g) + H₂O (l)

First we calculate how many H₂SO₄ moles reacted, using the given concentration and volume of solution:

moles = Molarity * Liters

• moles = 18.0 M * 2.50 L = 45 mol H₂SO₄

Then we convert H₂SO₄ moles into NaCO₂ moles, using the stoichiometric coefficients:

• 45 mol H₂SO₄ *
  `(1molNa_2CO_3)/(1molH_2SO_4)` = 45 mol NaCO₂

Finally we convert 45 NaCO₂ moles into grams, using its molar mass:

• 45 mol NaCO₂ * 106 g/mol = 4770 g