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

if one mole of calcium carbonate the limiting reactant is used how much calcium chloride should the reaction produce look at pic class=

Respuesta :

Answer:

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.