GrimLykReaper
New member
I remember learning this topic about 2 months ago but don't understand this question (although it's simple).
**A 30.0 cm3 sample of a 0.480 mol dm–3 solution of potassium hydroxide was partially neutralised by the addition of 18.0 cm3 of a 0.350 mol dm–3 solution of sulphuric acid.
(i)Calculate the initial number of moles of potassium hydroxide.
0.0144 moles
(ii)Calculate the number of moles of sulphuric acid added.
0.0063 moles
This is the part I don't quite get(iii)Calculate the number of moles of potassium hydroxide remaining in excess in the solution formed.
Based on the equation H2sO4 + 2KOH -> K2SO4 + 2H20 ;
I initially thought that the excess moles of KOH left over would've been 0.0144-0.0063 moles however my vague notes from class say that it is 0.0144-2(0.0063) moles
Is it that 2 moles of alkali have to neutralise 2 moles of acid.
Could someone just explain please because I can't seem to get my head around it!!
Thanks.
**A 30.0 cm3 sample of a 0.480 mol dm–3 solution of potassium hydroxide was partially neutralised by the addition of 18.0 cm3 of a 0.350 mol dm–3 solution of sulphuric acid.
(i)Calculate the initial number of moles of potassium hydroxide.
0.0144 moles
(ii)Calculate the number of moles of sulphuric acid added.
0.0063 moles
This is the part I don't quite get(iii)Calculate the number of moles of potassium hydroxide remaining in excess in the solution formed.
Based on the equation H2sO4 + 2KOH -> K2SO4 + 2H20 ;
I initially thought that the excess moles of KOH left over would've been 0.0144-0.0063 moles however my vague notes from class say that it is 0.0144-2(0.0063) moles
Is it that 2 moles of alkali have to neutralise 2 moles of acid.
Could someone just explain please because I can't seem to get my head around it!!
Thanks.