Question

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

Answer 1

The answer to your question is 111 g of CaCl₂

Step-by-step 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.