2022 hsc chem q (1 Viewer)

[indeed]

For part B, I'm confused of the two methods of solving for moles avaliable in the question:
the first is the way they show in the answers = using c times v
second way i thought of is getting mass of HCl and dividing by its MM, and doing same for NaOH, and finding limiting reagent

You end up with two different solutions so idk why the second way doesn't work

 

[Luukas.2]

The masses are given for the solutions but you also have their concentrations and volumes, from which you can see that the reagents are present in their stoichiometric ratio and neither is present in excess... so, no limiting reagent calculation is needed.

 

[Luukas.2]

So, you calculate in part (a), and then they give you a to use in part (b) to find . For part (b), it doesn't matter whether you use the moles of HCl or of NaOH as they are the same as the reaction is 1:1.

 

[indeed]

The masses are given for the solutions

You can still use n = m / MM for solutions right?

 

[Luukas.2]

You can still use n = m / MM for solutions right?

No, you can't. You know the mass of the whole solution (solute plus solvent) so you don't know the mass of NaOH or HCl present. You can only find the moles of each compound from n = CV.

 

[indeed]

No, you can't. You know the mass of the whole solution (solute plus solvent) so you don't know the mass of NaOH or HCl present. You can only find the moles of each compound from n = CV.

Oh ok I see, thank you so much!

 

[Unovan]

You can still use n = m / MM for solutions right?

no, because it is the mass of solution which includes the water as well

to find moles in hsc you either use n=m/MM or c=n/v