Acid spills are often neutralized with sodium carbonate. For example
Na2CO3 (s) + H2SO4 (aq) Na2SO4 (aq) + CO2 (g) + H2O (l)
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?
4.77 kg
How many moles do you have in the spilled H2SO4? That is
2.50L x 18.0 M = approximately 50 moles.
So you will need how many moles Na2CO3. Look at th equation. It's 1 mole Na2CO3 for 1 mol H2SO4; therefore, we will need about 50 (you need to do it more accurately.)
50 mols x molar mass Na2CO3 = grams
To determine the amount of sodium carbonate required to neutralize the spilled 2.50 L bottle of 18.0 M H2SO4, we need to use stoichiometry.
First, let's calculate the number of moles of H2SO4 in the spilled bottle:
Volume of H2SO4 = 2.50 L
Molarity of H2SO4 = 18.0 M
Moles of H2SO4 = Volume of H2SO4 x Molarity of H2SO4
= 2.50 L x 18.0 mol/L
= 45 mol
Next, since the balanced equation shows a 1:1 ratio between H2SO4 and Na2CO3, we can conclude that 45 moles of H2SO4 will require 45 moles of Na2CO3 to neutralize it.
Finally, let's calculate the mass of sodium carbonate required:
Molar mass of Na2CO3 = 2(Na) + 1(C) + 3(O)
= 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol)
= 46.00 g/mol + 12.01 g/mol + 48.00 g/mol
= 106.01 g/mol
Mass of Na2CO3 = Moles of Na2CO3 x Molar mass of Na2CO3
= 45 mol x 106.01 g/mol
≈ 4775 g
Therefore, approximately 4775 grams (or 4.775 kg) of sodium carbonate would be required to neutralize the spilled H2SO4.