When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO3(s)+2HCl(aq)⟶CaCl2(aq)+H2O(l)+ CO2(g) How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid? mass of CaCl2 :

Mass of CaCl₂ produced from the reaction = 21.31 g

Explanation:

The balanced equation for the reaction

CaCO₃(s) + 2HCl(aq) ⟶ CaCl₂(aq) + H₂O(l) + CO₂(g)

How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid

First of, we need to recognize the limiting reagent in this reaction.

The limiting reagent is the reagent that is totally used up in the chemical reaction; it determines the amount of other reactants that react and the amount of products that will be formed.

To know the limiting reagent, we convert the masses of the reactants given to number of moles.

Number of moles = (mass)/(molar mass)

25.0 g of calcium carbonate

Molar mass of CaCO₃ = 100.0869 g/mol

Number of moles of CaCO₃ present at the start of the reaction = (25/100.0869) = 0.250 moles

14.0 g of hydrochloric acid

Molar mass of HCl = 36.46 g/mol

Number of moles of HCl present at the start of the reaction = (14/36.45) = 0.384 moles

But from the stoichiometric balance of the reaction,

1 mole of CaCO₃ reacts with 2 moles of HCl

If CaCO₃ was the limiting reagent,

0.25 moles of CaCO₃ at the start of the reaction would require 0.50 moles of HCl; which is more than the available number of moles of HCl available at the start of the reaction (0.384 moles)

So, CaCO₃ isn't the limiting reagent.

If HCl is the limiting reagent,

0.384 moles of HCl would require 0.192 moles of CaCO₃ to react with. This is within the limit of CaCO₃ present at the start of the reaction. (0.250 moles)

Hence, HCl is the limiting reagent, it is the reactant that is used up in the reaction and determines the amount of products formed.

Again, from the stoichiometric balance of the reaction,

2 moles of HCl gives 1 mole of CaCl₂

0.384 moles of HCl will give (0.384/2) moles of CaCl₂; that is, 0.192 moles of CaCl₂

We then convert this number of moles to mass.

Mass = (number of moles) × (molar mass)

Molar mass of CaCl₂ = 110.98 g/mol

Mass of CaCl₂ produced by the reaction = 0.192 × 110.98 = 21.30816 g = 21.31 g

Hope this Helps!!!

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the answer is na (sodium)