[Florence]

H₂SO₄ + 2NaOH Na₂SO₄ + 2H₂O

I used 27.6cm³ of a 0.5M solution of NaOH to neutralize 25cm³ of H₂SO₄

My actual yield of Na₂SO₄ crystals was 0.7g.

I need to work out the percentage yield but I'm stuck on how to work out the theoretical yield from the above data (or do I need to give more). Please help,

Thanks,
Florence

[Schrödinger]

H₂SO₄ + 2NaOH Na₂SO₄ + 2H₂O

I used 27.6cm³ of a 0.5M solution of NaOH to neutralize 25cm³ of H₂SO₄

Isn't the concentration of H₂SO₄ given?

[Florence]

H₂SO₄ + 2NaOH Na₂SO₄ + 2H₂O

I used 27.6cm³ of a 0.5M solution of NaOH to neutralize 25cm³ of H₂SO₄

Isn't the concentration of H₂SO₄ given?

I think it was one of things we had to work out and then use to work out the theoretical yield. I tried working it out but I'm not sure if it's correct.

I did (0.5/1000) x 27.6 to work out the amount of moles of NaOH used in the reaction which gave me 0.0138mol.

As 2 moles of NaOH is needed to neutralize 1 mole of H₂SO₄ I then did 0.0138/2 to work out the amount of moles of H₂SO₄ produced. Which gave me 0.0069mol. So now I know that in 25cm³ of H₂SO₄ there is 0.0069mol so I worked out the molar concentration (0.0069/25) x 1000 = 0.276M = Molar concentration of H₂SO₄

I'm not sure if this was the correct way to go about it, or how to use this to work out the theoretical yield though :S

[Schrödinger]

Okay. When you said 'to neutralize' I thought the reaction was aimed at neutralizing the acid, but wasn't a complete neutralization (i.e., there was a limiting reactant). In the case that the acid is COMPLETELY neutralized, what you have done is correct. You indeed do calculate the concentration of the acid from the given data.

Now that you know that the reaction is quantitative, moles of Na₂SO₄ produced theoretically is the 0.0069 (same as that of H₂SO₄, as can be seen from the stoichiometry of the equation). Find out the theoretical mass of  Na₂SO₄ and hence the yield.

[Florence]

Okay. When you said 'to neutralize' I thought the reaction was aimed at neutralizing the acid, but wasn't a complete neutralization (i.e., there was a limiting reactant). In the case that the acid is COMPLETELY neutralized, what you have done is correct. You indeed do calculate the concentration of the acid from the given data.

Now that you know that the reaction is quantitative, moles of Na₂SO₄ produced theoretically is the 0.0069 (same as that of H₂SO₄, as can be seen from the stoichiometry of the equation). Find out the theoretical mass of  Na₂SO₄ and hence the yield.

So, the Relative Molar Mass of Na₂SO₄ is 142g

Would I do (142/1) x 0.0069 = Theoretical yield of Na₂SO₄ ?

[Schrödinger]

So, the Relative Molar Mass of Na₂SO₄ is 142g

Would I do (142/1) x 0.0069 = Theoretical yield of Na₂SO₄ ?

Yes. That's correct

[Florence]

So, the Relative Molar Mass of Na₂SO₄ is 142g

Would I do (142/1) x 0.0069 = Theoretical yield of Na₂SO₄ ?

Yes. That's correct

Yayyy! Thank you so much!