Chemistry Limiting Reagent

Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO₃ + 2HCl  CaCl₂ + H₂O + CO₂

How many grams of calcium chloride will be produced when 25g of calcium carbonate is combined with 14g of hydrochloric acid?

How many grams of the excess reactant will remain after the reaction is complete?

mass of excess reactant:

Answer 1

First step here is to determine the limiting reactant. Your equation is balanced and shows that for every mole of CaCO₃ you need 2 moles of HCl. So determine the moles of CaCO₃ and HCl by dividing the amount of each one by its formula weight.

moles CaCO₃ = 25 g/100 g/mole = 0.25 mole

moles HCl = 14g/36.5 g/mole = 0.38 mole

You need 0.5 mole of HCl to react with the 0.25 mol of CaCO₃ , but you have less than that. So the HCl is the limiting reactant and CaCO₃ is the excess.

The moles of product, CaCl₂ , will be half the number of moles HCl used, according to your equation. So that means from 0.38 moles of HCl, you can produce 0.19 moles of CaCl₂.

Therefore grams of CaCl₂ produced will be (0.19 g)x(111 g/mole CaCl₂) = 21 g

Moles of CaCO₃ needed to react completely with 0.38 mole HCl will be 0.19 because you need 2 moles of HCl to react with every mole of CaCO₃.

Number of moles of CaCO₃ used up in this reaction will be 0.25 - 0.19 = .06 moles leftover.

Simply multiply that by the molar mass of CaCO₃, which is 100 g/mole: (0.06 moles)x(100 g/mole) = 6 grams of excess reactant left after the reaction.