Hydrchloric acids

Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. 

CaCO3 + 2HCl = CaCl2 (aq) + H2O + CO2
A.)How many grams of calcium chloride will be produced when 32.0 g of calcium carbonate are combined with 14.0 g of hydrochloric acid?
B.)Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

Answer 1

For such questions, look at the BALANCED EQUATION.  It tells you the mole ratio of all reaction and products.

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

1 mole CaCO₃ reacts with 2 moles HCl to produce 1 mole CaCl₂, 1 mole H₂O and 1 mole CO₂.

(A) Limiting reactant: Which reactant will run out first?

moles of CaCO₃ used = 32.0 g x 1 mole/100g = 0.320 moles

moles HCl used = 14.0 g x 1 mole/36.45 g = 0.384 moles.  BUT since it takes 2 moles HCl for every 1 moles CaCO₃, HCl is limiting. It will run out first.  Thus, the moles of HCl will dictate how many moles of CaCl2 can be produced.

Moles of CaCl₂ produced = 0.384 moles HCl x 1 mole CaCl₂/2 moles HCl = 0.192 moles

Grams CaCl2 produced = 0.192 moles x 110.98 g/mole = 21.31 grams = 21.3 g (to 3 significant figures)

 

(B)  Limiting reactant is HCl as described above, which makes CaCO3 in excess.  We can determine how much is left over as follows.  Find number of moles used up in the reaction and subtract it from the number of moles at the start.  That gives us moles left over.  Convert this to grams using molar mass.

moles CaCO₃ used = 0.384 moles HCl x 1 mole CaCO₃/2 moles HCl = 0.192 moles used

moles CaCO₃ left = 0.320 moles - 0.192 moles = 0.128 moles

grams CaCO₃ left = 0.128 moles x 100 g/mole = 12.8 grams left over.