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

Question

Peyton Adams · Accepted Answer

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

The stoichiometric equation is required; #CaCO_3(s) + Delta rarr CaO(s) + CO_2(g)uarr# The quantitative reaction shows that one mole of carbon dioxide is evolved for every mole of calcium carbonate, but you have to heat this fairly intensely. #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)# When carbon dioxide is added to filtered limewater, calcium carbonate is reformed and separates out of the 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.

Natalie Anton · Answer

undefined The molar mass of CO2 is 44.01 g/mol. Therefore, the decomposition of 3.00 mol of CaCO3 will produce $3.00 \text{ mol} \times 44.01 \text{ g/mol} = 132.03 \text{ g}$ of CO2.