# Question #e2c57

Dec 20, 2016

Here's what I got.

#### Explanation:

You cannot determine the limiting reagent without a balanced chemical equation, so make sure that you have that part covered before moving on.

${\text{CaCl"_ (2(aq)) + "K"_ 2"CO"_ (3(aq)) -> "CaCO"_ (3(s)) darr + 2"KCl}}_{\left(a q\right)}$

Notice that you have a $1 : 1$ mole ratio between the two reactants; this tells you that the reaction consumes calcium chloride and potassium carbonate at the same rate.

Simply put, for every mole of calcium chloride that takes part in the reaction, the reaction consumes $1$ mole of potassium carbonate.

Now, you're mixing solutions of equal volumes and equal molarities, so right from the start you know that you're mixing equal numbers of moles of each reactant.

As you know, the limiting reagent is the reactant that gets consumed first in a chemical reaction. In other words, the reactant that gets consumed before all the moles of the second reactant get the chance to take part in the reaction will be your limiting reagent.

In this case, you have equal numbers of moles of calcium chloride and potassium carbonate, so which one will be consumed first?

The answer is that the two reactants get consumed at the same time because of the $1 : 1$ mole ratio and of the fact that you have equal numbers of both available.

You can thus say that neither calcium chloride, nor potassium carbonate will act as a limiting reagent.