Question

If 3.00 mol of `CaCO_3` undergo decomposition from `CaO` and `CO_2` how many grams of `CO_2` are produced?

Answer 1

Clearly, `3.00` `mol` of carbon dioxide gas are evolved.

Explanation:

We need the stoichiometric equation;

`CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr`

You have to heat this fairly fiercely, but clearly, quantitative reaction indicates that 1 mol of carbon dioxide is evolved per mole of calcium carbonate.

`CaO` dissolves in water to give limewater, aqueous calcium hydroxide (this is sparingly soluble, so filter it);

`CaO(s) + H_2O(l) rarr Ca(OH)_2(aq)`

If filtered limewater is treated with carbon dioxide, calcium carbonate is reformed, and precipitates from solution as a milky white solid:

`Ca(OH)_2(s) + CO_2(g) rarr CaCO_3(s)darr + H_2O(l)`

All of these equations should be informed by doing the actual reactions in the laboratory. The best source of `CO_2` is sparkling water, the type you would drink for lunch; this is supersaturated with respect to carbon dioxide.