#### Archived

This topic is now archived and is closed to further replies.

# Chem Equation help?!

Hi!
I'm very stuck on a chemistry equation, after having done a back titration.

Here's what I have;

Mass Calcium Carbonate used: 2g
Accurate volume NaOH; 4.5 cm3
Amount of HCl used: 40 ml

HCl solution made up to 200 ml with distilled water

CaCO3 + 3HCl = CaCl2 + CO2 + H20

NaOH + HCl = H20 + NaCl

Moles NaOH used: 0.00045
Moles HCl in 20 cm3 of mixture; 0.00045 moles

Moles HCl in 200 cm3 of mixture: 0.0045

Moles HCl originally added: 0.08 moles

Amount HCl used up: 0.0755

Moles CaCO3: 0.03775

1 Mole CaCO3: 100g

Mass CaCO3 reacted: 3.775 g

% purity= 188.75

I know that it can't be 188.75 % pure. So what I want to know, is where I'm going wrong. If anyone can help, thanks in advance! It's for my EE, so I really need to know why my result is so off-course.

##### Share on other sites

Moles HCl in 20 cm3 of mixture; 0.00045 moles

Moles HCl in 200 cm3 of mixture: 0.0045

I think you might have an error in this part. Correct me if I'm wrong as it's been a while since I did Chemistry but I don't think you can have it MORE concentrated in 200cm3 than it was in 20cm3 if you made it up to 200ml with distilled water. It must be the other way around.

1 person likes this

##### Share on other sites

Here's what it says in my chem equation book;

From the equation, you need the same number of moles of HCl and NaOH (so, same number of moles as NaOH, = 0.00045_

This was a 25.0cm3 sample from the whole 250cm3 of the reaction mixture.

Therfore, number of moles of HCL laft over in the wohle mixture after reacting with the limestone = 10x0.003 (This is the example from the book, not my example) = 0.03000 mol

This is also what my chem teacher told me to do.