wings Posted June 1, 2013 Report Share Posted June 1, 2013 Hi!I'm very stuck on a chemistry equation, after having done a back titration.Here's what I have;Mass Calcium Carbonate used: 2gAccurate volume NaOH; 4.5 cm3Amount of HCl used: 40 mlHCl solution made up to 200 ml with distilled waterCaCO3 + 3HCl = CaCl2 + CO2 + H20NaOH + HCl = H20 + NaClMoles NaOH used: 0.00045Moles HCl in 20 cm3 of mixture; 0.00045 molesMoles HCl in 200 cm3 of mixture: 0.0045Moles HCl originally added: 0.08 molesAmount HCl used up: 0.0755Moles CaCO3: 0.037751 Mole CaCO3: 100gMass CaCO3 reacted: 3.775 g% purity= 188.75I 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.Thanks in advance! Reply Link to post Share on other sites More sharing options...
Sandwich Posted June 1, 2013 Report Share Posted June 1, 2013 Moles HCl in 20 cm3 of mixture; 0.00045 molesMoles HCl in 200 cm3 of mixture: 0.0045I 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 Reply Link to post Share on other sites More sharing options...
wings Posted June 1, 2013 Author Report Share Posted June 1, 2013 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 molThis is also what my chem teacher told me to do. Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.