Question

Given a `20*g` mass of calcium carbonate, how much carbon dioxide will be generated, when it is added to a `100*mL` volume of `"20% w/w"` solution of `HCl(aq)`?

Answer 1

The units of concentration of the hydrochloric acid are unfortunate.....I gets approx. a `5*L` volume under standard conditions.

Explanation:

We interrogate the reaction......

`CaCO_3(s) + 2HCl(aq) rarr CaCl_2(aq) + H_2O(l) + CO_2(g)uarr`

Now this site reports that the density of `20%` `HCl` is `rho=1.098*g*mL^-1`. These data SHOULD have been supplied with the question.

And thus moles of `HCl=(20%xx100*cancel(mL)xx1.098*cancelg*cancel(mL^-1))/(36.46*cancelg*mol^-1)`

`=0.602*1/(mol^-1)=0.602*1/(1/(mol))=0.602*mol`

And thus the molar concentration of the hydrochloric acid was approx. `6*mol*L^-1`.

And also moles of `"calcium carbonate"-=(20*g)/(100.09*g*mol^-1)`

`=0.200*mol`.

There is thus a stoichiometric excess of hydrochloric acid, and calcium carbonate is the limiting reagent. A `0.200*mol` quantity of carbon dioxide gas will be released.

Depending on your syllabus, we know that the molar volume at `1*atm` and `298*K` is `24.5*L*mol^-1`. You have to check with the given definitions in your course.

And thus `"volume"=24.5*L*mol^-1xx0.200*mol=??*L.`

How much acid is in excess?