CaCO\(_3\) → CaO + CO\(_2\)
CaCO\(_3\) = 40 + 12 + (16 x 3) = 100,

CaO = 40 + 16 = 56
 56g of CaO was produced by 100g of CaCO\(_3\)
∴ 5.6g of CaO will be produced by
(5.6 x 100)/56 = 10g of CaCO\(_3\)

Recall that Number of moles = \(\frac{mass}{molar mass}\)

  = \(\frac{10}{100}\)

  = 0.10mole

Alternatively,

 CaCO\(_3\)  → CaO + CO\(_2\)

  n = \(\frac{m}{M}\)

  n = \(\frac{5.6}{56}\)

  n = 0.10 mole

Since the ratio of the reactant to product is 1 : 1, and the number of mole of CaO is 0.10mole, therefore, the number of mole of CaCO\(_3\) will also be 0.10 mole, according to Avogadro's law.