Question

How many atoms of oxygen are present in 300 grams of CaCO3? atomic mass -Ca=40u, C=12u,) = 16 u

Answer 1

Molecular mass of calcium carbonate = 40 + 12 + 48 =100g

1 mole of Calcium carbonate contains = 3 mole of oxygen atoms

100 g of Calcium carbonate contains = 3 X 6.022 X 10∧23 oxygen atoms

300g of Calcium carbonate contains How many oxygen atoms?

= 300 X 3 X 6.022 X 10∧23/ 100 = 54.205 X 10∧23 oxygen atoms

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