CaCO3 + HCl

See diagram. 'A student carried out an experiment to investigate the rate of reaction between powdered calcium carbonate and excess dilute hydrochloric acid, using the following apparatus: (diagram). (N.B the grey in the neck of the flask is cotton wool)

a) The student measured the mass of the flask and contents every 30 seconds and calculated the loss in mass:

Time (min) | Loss of mass (g)
----------------------------
***0.0***|***0.0***
***0.5***|***0.6***
***1.0***|***1.05**
***1.5***|**1.45***
...etc!

(i) Explain why there is a loss in mass.
(ii) Suggest why the student used a loose plug of cotton wool in the neck of the flask.

Can someone help me with this please!
Thanks

ah ok thanks. I didn't think the gas was heavy enough to make any significant mass change.

just a thought; if mass loss from the gas was ignored, and all the solid CaCO3 turned into CaCl2 so that there was just a mixture of CaCl2 (aq) and HCl, would there be any change in mass?

Quote:
Originally Posted by Lozza
you cant ignore the loss of gas!!

how on earth could it just produce one product?!?!?!


I dunno! im confused sorry ignore me!

I'll set the record straight:

CaCO3 + 2HCl -> CO2 + H2O + CaCl2

The loss in mass is caused by the loss of CO2. CO2 does have mass, it's just that gas has a low density (takes up a lot of room, more than an equivalent mass of solid), but a large volume of it is given off in this reaction.

For 100g (=1 mole) of CaCO3 which reacts with 73g (=2 moles) of HCl:
44g becomes CO2
18g becomes H2O
111g becomes CaCl2

The cotton wool is used to prevent the contents from evaporating or leaving the contained, but allowing the CO2 to pass through. Using a normal bung, the pressure inside the container (due to the gas being produced) would make the bung fly out.

Hope that helps!

Ferg