if you wanted to create 500g of calcium carbonate how much calcium chloride and sodium carbonate would you need?I need help please anyone. It's using stiochonometry and limiting reagent both or either one I don't know. But please help me​

Mass of Calcium Carbonate = 500 grams

The required reaction would be:

CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl

Number of moles of CaCO₃:

Number of moles = given mass / molar mass

Number of moles = 500 / 100              [molar mass of CaCO₃ = 100g/mol]

Number of moles = 5 moles

Using the power of stoichiometry!!!

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

it is called acid bonding