Mass of Calcium Carbonate = 500 grams
The required reaction would be:
CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl
<u>Number of moles of CaCO₃:</u>
Number of moles = given mass / molar mass
Number of moles = 500 / 100 [molar mass of CaCO₃ = 100g/mol]
Number of moles = 5 moles
<u>Using the power of stoichiometry!!!</u>
CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl
from the balanced reaction listed above, we can see that one mole of CaCl₂ and one mole of Na₂CO₃ produces one mole of CaCO₃
Moles of CaCl₂ = Moles of Na₂CO₃ = Moles of CaCO₃
we just solved for the number of moles of CaCO₃, which gives us
Moles of CaCl₂ = Moles of Na₂CO₃ = 5 moles
Hence, 5 moles of CaCl₂ and Na₂CO₃ each is required
