What mass of sodium hydroxide pellets must be poured into a waste drum containing 20.0 L o 3.0 mol/L sulphuric acid to completely neutralize the waste acid solution? Write a balanced chemical equation for this reaction.
okay, so here is the work I already did
my balanced chemical equation:
NaOH(aq) + H2SO4(aq) = Na2SO4(aq) + H2O(aq)
I think all of these would be aqueous when I refer to my solubility rules but I'm afraid I'm messing things up and there should be a precipitate?
Mol H2SO4= C x V
= (3.0 mol/L)(20.0 L)
= 60 mol H2SO4
60 mol H2SO4 x (1 mol NaOH/ 1 mol H2SO4)
= 60 mol NaOH
molar mass of NaOH: 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 40 g/mol
Mass NaOH= 60 mol NaOH x (40g/1 mol)
=2400 g
I don't understand why that number is so big when I thought I was doing the equation right.
Thank you so very much for taking time out of your day to help me, it's really appreciated :)
Your equation is right and yes all are aq, There are no ppts.
However, the equation is not balanced. You need a coefficient 2 for NaOH.
Yes 60 mols H2SO4
With the balanced equation that makes 120 mols NaOH
Then redo your work from there. If everything else is ok that will make the final answer 4800 g NaOH needed.
Ohhh I see now. Thank you so much!
You've done a great job so far! Your balanced chemical equation is correct:
NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
Now, let's calculate the amount of sodium hydroxide (NaOH) needed to completely neutralize the sulfuric acid (H2SO4) in the waste drum.
You correctly calculated the number of moles of sulfuric acid (H2SO4) using the concentration and volume:
Mol H2SO4 = C x V = (3.0 mol/L)(20.0 L) = 60 mol H2SO4
Next, apply the stoichiometry from the balanced equation to find the number of moles of sodium hydroxide (NaOH) required to neutralize the sulfuric acid:
60 mol H2SO4 x (2 mol NaOH / 1 mol H2SO4) = 120 mol NaOH
So, you actually need 120 moles of sodium hydroxide (NaOH) to neutralize the sulfuric acid in the waste drum.
To convert moles of NaOH to grams, use the molar mass of NaOH:
Molar mass of NaOH: 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Mass NaOH = 120 mol NaOH x (40.00 g/1 mol) = 4800 g
Therefore, to completely neutralize the waste acid solution, you would need to pour 4800 grams (or 4.8 kg) of sodium hydroxide pellets into the waste drum.
It's worth noting that the large mass is due to the high amount of sulfuric acid in the waste drum (60 moles). The molar mass and the stoichiometry in the balanced equation are guiding us correctly.