# How many oxygen atoms would it produce if it is completely decomposed? Consider a sample of calcium carbonate in the form of a cube measuring 1.25 in on each edge. If the sample has a density of...

**How many oxygen atoms would it produce if it is completely decomposed? **

Consider a sample of calcium carbonate in the form of a cube measuring 1.25 in on each edge. If the sample has a density of 2.71 g/cm3, how many oxygen atoms would it produce if it is completely decomposed?

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Calcium carbonate is CaCO3. If heated to around 800 degrees C, it thermally decomposes to CaO and CO2. The equation is written below:

CaCO3 --> CaO + CO2

If by "how many oxygen atoms would it produce" you mean how many oxygen atoms would be liberated as a part of CO2, there there will be 2 atoms of O produced per molecule of CaCO3 burned. So we need to figure out the moles of CaCO3 in the solid sample. The cube has a side length of 1.25 inches. Since there are 2.54 cm per inch, this gives:

1.25 in (2.54 cm/1 in) = 3.175 cm

To find the volume of the cube, simply cube the length of one side:

(3.175 cm)^3 = 32.006 cm^3

Multiply the volume by the density to get the mass of the sample in grams:

32.006 cm^3 (2.71 g/cm^3) = 86.74 g CaCO3

Now divide by the molecular weight of CaCO3 to get the moles:

86.74 g (1 mole/100.09 g) = 0.867 moles CaCO3

Every mole of CaCO3 gives one mole of CO2 which contains 2 moles of oxygen for a total of 0.867*2 = 1.734 moles O. Now multiply by Avogadro's number to get the total number of atoms of oxygen:

1.734 moles O (6.022 x 10^23 atoms/1 mole) = 1.04 x 10^24 atoms O.

So a total of 1.04 x 10^24 oxygen atoms would be produced.

If by totally decomposed you mean that all three atoms of oxygen are somehow released, then multiply the moles of CaCO3 times 3 and then multiply by Avogadro's number:

2.601 moles O (6.022 x 10^23 atoms/1 mole) = 1.57 x 10^24 atoms O.