When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO 3 ( s ) + 2 HCl ( aq ) ⟶ CaCl 2 ( aq ) + H 2 O ( l ) + CO 2 ( g ) How many grams of calcium chloride will be produced when 28.0 g of calcium carbonate is combined with 10.0 g of hydrochloric acid? mass of CaCl 2 : 110.98 g Which reactant is in excess? CaCO 3 HCl How many grams of the excess reactant will remain after the reaction is complete? mass of excess reactant:

Mass of formed CaCl₂ is 14.9 g

Mass of CaCO₃ that remains after the reaction is 14.5 g

Explanation:

We determine the reaction:

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

We have the mass of both reactants, so we have to find out which is the limiting; firstly we convert the mass to moles

28 g . 1mol / 100g = 0.28 moles of CaCO₃

10 g . 1mol /36.45g =0.27 moles of HCl

Ratio is 1:2. Let's make a rule of three:

1 mol of carbonate needs 2 moles of acid, to react

Then, 0.28 moles must react with the double of moles, of acid.

Therefore the HCl is the limiting reactant, and the reagent in excess is the salt.

Ratio is 2:1. The rule of three will be:

2 moles of acid needs 1 mol of salt to react

Then, 0.27 moles of acid, will react with (0.27 .1 ) / 2 = 0.135 moles of salt

We have 0.28 moles of carbonate so (0.28 - 0.135) = 0.145 moles of salt will remain after the reaction is complete.

We convert the moles to mass → 0.145 mol . 100 g/1mol = 14.5 g

We said, that the limiting reagent was the HCl so we can work with the products, now. Ratio is 2:1

2 moles of HCl can produce 1 mol of CaCl₂

Then, 0.27 moles of HCl will produce (0.27 . 1)/ 2 = 0.135 moles of chloride

We convert the moles to mass → 0.135 mol . 110.98 g/1mol = 14.9 g

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