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

Mar 22, 2016

Clearly, $3.00$ $m o l$ of carbon dioxide gas are evolved.

Explanation:

We need the stoichiometric equation;

$C a C {O}_{3} \left(s\right) + \Delta \rightarrow C a O \left(s\right) + C {O}_{2} \left(g\right) \uparrow$

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.

$C a O$ dissolves in water to give limewater, aqueous calcium hydroxide (this is sparingly soluble, so filter it);

$C a O \left(s\right) + {H}_{2} O \left(l\right) \rightarrow C a {\left(O H\right)}_{2} \left(a q\right)$

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

$C a {\left(O H\right)}_{2} \left(s\right) + C {O}_{2} \left(g\right) \rightarrow C a C {O}_{3} \left(s\right) \downarrow + {H}_{2} O \left(l\right)$

All of these equations should be informed by doing the actual reactions in the laboratory. The best source of $C {O}_{2}$ is sparkling water, the type you would drink for lunch; this is supersaturated with respect to carbon dioxide.