If one mole of calcium carbonate (the limiting reactant) is used, how much calcium chloride should the reaction produce? look at pic

The answer to your question is 111 g of CaCl₂

Explanation:

Reaction

                 2HCl  +  CaCO₃   ⇒    CaCl₂  +  CO₂  +  H₂O

Process

1.- Calculate the molecular mass of Calcium carbonate and calcium chloride

CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g

CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g

2.- Calculate the amount of calcium chloride produced using proportions.

The proportion CaCO₃ to CaCl₂ is   1 : 1.

                      100 g of CaCO₃   111 g of CaCl₂

Then 111g of CaCl₂ will be produced.

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