Calcium carbonate decomposes at high temperatures forming calcium oxide and carbon dioxide. How much carbon dioxide is produced if 14.75 g of...

The decomposition reaction of calcium carbonate is given as:

`CaCO_3 (s) + Heat -> CaO (s) + CO_2 (g)`

Here, calcium carbonate decomposes to calcium oxide and carbon dioxide.

Using stoichiometry, 1 mole of calcium carbonate produces 1 mole of calcium oxide and 1 mole of carbon dioxide.

First we need to calculate the molar mass of all the compounds. The molar mass of calcium carbonate is 100 g (= 40 + 12 + 3 x 16), of calcium oxide is 56 g (= 40 + 16) and that of carbon dioxide is 44 g (= 12 + 2 x 16).

Thus, 100 g of calcium carbonate decomposes to produce 56 g of calcium oxide and 44 g of carbon dioxide. 

Here, 14.75 g of calcium carbonate is decomposed.

We will first calculate how many moles of calcium carbonate is decomposed:

14.75 g CaCO3 x (1 mole CaCO3 / 100 g CaCO3) = .1475 moles CaCO3 decomposed

We know that the molar ratio of CaCO3 decomposed and CO2 produced is 1:1. Therefore, .1475 moles of CO2 will be produced.

To put this into grams:

.1475 moles CO2 x (44 grams CO2 / 1 mole CO2) = 6.49 grams of CO2 produced

Thus, 6.49 g of carbon dioxide is produced when 14.75 g calcium carbonate decomposes.

Hope this helps.