# Homework Help: Confused with moles

1. Nov 24, 2007

### Trail_Builder

I thought I knew moles inside out, but I havnt quite been told how to factor in what happens when you dissolve something in something.

1. The problem statement, all variables and given/known data

1.500g of a sample of limestone was dissolved in 50cm^3 of a 1.00 mol/dm^3 of hydrochloric acid solution. The resulting solution was made up to 250cm^3 precisely with distilled water. 25.0cm^3 of this solution required 21.05cm^3 of 0.100 mol/dm^3 of sodium hydroxide for neutralisation.

Calculate the % of rock which is calcium carbonate.

2. Relevant equations

3. The attempt at a solution

mol of HCL:

mol = (vol/1000)*conc
50cm^3 is 1 mol/dm^3, so added with water up to 250cm^3 will be 0.2 mol/dm^3

mol(HCl) = 0.250*0.2
mol(HCl) = 0.05

but then we only take 25cm^3, so...

mol(HCl) = 0.005

mol of NaOH:

mol(NaOH) = (vol/1000)*conc
mol(NaOH) = (21.05/1000)*0.1

mol(NaOH) = 0.002105

Now I get stuff, Im not sure what the effect of dissolving 1.5g of limestone will even have, so Im stuck :S.

Hope you can help.

Thanks

Last edited: Nov 24, 2007
2. Nov 24, 2007

### mgb_phys

The NaOH is needed to neutralise any HCl that wasn't 'used up' by the limestone - you just have to work backwards.

1, Work out many moles of NaOH
2, How many moles of HCl would this react with
3, How much HCl was made up originally.
4, - subtract (2)
5, how much CaCO3 would this HCl react with?

3. Nov 24, 2007

### Trail_Builder

so for 1) I have 0.002105
for 2) 0.002105 (because 1:1 ratio)
3) 0.005
4) 0.002895
5) 0.0014475 mol of CaC03 right? (because the equation would be CaCO3 + 2HCl goes to CaCl2 + H2O + CO2, so 1:2 ratio)

just checking lol.