# Alevel Chemistry - Moles

12.0g of a mixture of calcium carbonate and sodium chloride was treated with 100cm3 of 2.00moldm-3 hydrochloric acid (only the calcium carbonate reacts). The resulting solution was made from 250cm3 with water and a 25.0cm3 portion of this needed 34.1cm3 of 0.200moldm-3 sodium hydroxide for neutralisation. Find the % by mass of the calcium carbonate in the mixture.

Hi, if someone could please help me with this question. I don't understand why you have to calculate the moles of HCl twice, once by using the volume and concentration and the next with the moles being the same with the NaOH (mole ratio) and then taking these two values away from each other? How would I know to calculate the moles twice?
(edited 1 year ago)
Original post by charlotte05x
12.0g of a mixture of calcium carbonate and sodium chloride was treated with 100cm3 of 2.00moldm-3 hydrochloric acid (only the calcium carbonate reacts). The resulting solution was made from 250cm3 with water and a 25.0cm3 portion of this needed 34.1cm3 of 0.200moldm-3 sodium hydroxide for neutralisation. Find the % by mass of the calcium carbonate in the mixture.

Hi, if someone could please help me with this question. I don't understand why you have to calculate the moles of HCl twice, once by using the volume and concentration and the next with the moles being the same with the NaOH (mole ratio) and then taking these two values away from each other? How would I know to calculate the moles twice?

Ah, this is what’s called a back titration.

It works by reacting what you have with a known excess of an acid, causing some of the acid to be neutralised and you then use a titration to work out how much of the acid is left, so that you can work out how much reacted and therefore how much of the substance there was to begin with.

So moles of HCl using titration data = moles of HCl not used initially

Moles of HCl added at the very start - moles of HCl not used initially = moles of HCl that reacted in the first reaction
Original post by TypicalNerd
Ah, this is what’s called a back titration.

It works by reacting what you have with a known excess of an acid, causing some of the acid to be neutralised and you then use a titration to work out how much of the acid is left, so that you can work out how much reacted and therefore how much of the substance there was to begin with.

So moles of HCl using titration data = moles of HCl not used initially

Moles of HCl added at the very start - moles of HCl not used initially = moles of HCl that reacted in the first reaction

Oh, that makes sense, I don’t think my teacher has gone over back titrations. However, how do you know that the acid is in excess and didn’t react at the start? Thank you sm !!
Original post by charlotte05x
Oh, that makes sense, I don’t think my teacher has gone over back titrations. However, how do you know that the acid is in excess and didn’t react at the start? Thank you sm !!

They will always choose an amount of acid that would be more than enough to react with the mixture, assuming it is 100% CaCO3. That will leave some acid unreacted whist a fair proportion of it will have been used.
Original post by TypicalNerd
They will always choose an amount of acid that would be more than enough to react with the mixture, assuming it is 100% CaCO3. That will leave some acid unreacted whist a fair proportion of it will have been used.

Ohhhh, okay so because in the question it said only the calcium carbonate reacted that means the acid is in excess?
Original post by charlotte05x
Ohhhh, okay so because in the question it said only the calcium carbonate reacted that means the acid is in excess?

I guess that is a fair way of thinking about it.