Is there always a limiting reagent in every equation? For instance, is there a limiting reagent in $\ce{CaCl2 + Na2CO3 -> 2NaCl + CaCO3}$ where the same number of moles are reacting with each other?

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    $\begingroup$ Considering that there are $6.022\times10^{23}$ molecules per mole, it seems unlikely that an exactly 1:1 ratio could ever really be achieved. $\endgroup$ – MaxW May 3 '18 at 16:36
  • $\begingroup$ @MaxW It is possible using e.g. AFM tips to nudge individual atoms and molecules and cause them to react, which I suppose counts as exact in a very technical manner. $\endgroup$ – Nicolau Saker Neto May 4 '18 at 13:26

An interesting philosophical question. We are given a chemical equation: $$\ce{CaCl2 + Na2CO3 -> 2NaCl + CaCO3}$$ and that the same number of moles of the reactants are used. Now the OP poses the question "Is there a limiting reagent?"

As chemists we can say that 1.00 moles of $\ce{CaCl2}$ were reacted with 1.00 moles of $\ce{Na2CO3}$. We'd pragmatically say that the moles were equal and that there was no limiting regent.

However this is a significant figure argument, not reality. Avogadro's number is about $6.022\times10^{23}$. There is no lab in the world that can weight out precisely one mole of $\ce{CaCl2}$ accurate to 23 decimal places. We don't even know Avogadro's number accurate to 23 decimal places! The best value is $6.022140857(74)\times10^{23}$.

So in reality, despite our best efforts as chemists, one reagent or the other will be in some small excess which we'd consider to be insignificant given the overall precision of the experiment.

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    $\begingroup$ Note that, since 20 May 2019, the Avogadro number is (by definition) exactly $6.022\,140\,76\times10^{23}$. $\endgroup$ – Loong Sep 7 at 20:48
  1. There can't be any limiting reagents in the equations. Equations are purely theoretical expressions and are always balanced in terms of moles.

  2. "Limiting reagents" arise in real world chemical reactions.

The questions on limiting reagents are usually formulated in a form of "we have a reaction according to this equation, and we have X grams of this reagent and Y grams of that reagent, which one is limiting?"

A question "which reagent in this equation is limiting?" without additional info makes no sense.

  • $\begingroup$ The OP has clearly stated "same number of moles are reacting with each other", so the amount of the reagents are given. Using the molar masses you can get to the actual weight. What additional info is required? $\endgroup$ – Gaurang Tandon May 4 '18 at 12:15

I'd like to explain this problem with an example. The chemical equation in question is:$$\ce{CaCl2 + Na2CO3 -> 2NaCl + CaCO3}$$ Since $K_\mathrm{sp}$ of $\ce{CaCO3}$ is $\pu{3.36\times10^{-9} mol^2\cdot L^{-2}}$, it is safe to assume that this reaction would have proceeded to completion, if $\pu{1 mol}$ of $\ce{CaCl2}$ is reacted with $\pu{1 mol}$ of $\ce{Na2CO3}$ to give $\pu{1 mol}$ of $\ce{CaCO3}$ solid and $\pu{2 mol}$ of aqueous $\ce{NaCl}$. Therefore, no limiting reagent in that calculation because we used exact stoichiometric amounts of starting materials.

What if you have mixed $\pu{100 mL}$ of $\pu{2 M}$ $\ce{CaCl2}$ with $\pu{100 mL}$ of of $\pu{1 M}$ $\ce{Na2CO3}$? You can calculate how many mols in each solution:

$$ \begin{align} \text{Number of moles of}~ \ce{CaCl2}~ \text{in solution} &= M_1V_1=2~\pu{mol L-1}\cdot \pu{0.100L}=\pu{0.2mol}\\ \text{Number of moles of}~ \ce{Na2CO3}~ \text{in solution} &= M_2V_2=1~\pu{mol L-1}\cdot \pu{0.100L}=\pu{0.1mol} \end{align}$$

Now, according to the chemical equation, only $\pu{0.1 mol}$ of $\ce{CaCl2}$ would react with available $\pu{0.1 mol}$ of $\ce{Na2CO3}$ to give $\pu{0.1 mol}$ of $\ce{CaCO3}$ solid and $\pu{0.2 mol}$ of aqueous $\ce{NaCl}$. The rest of $\pu{0.1 mol}$ of $\ce{CaCl2}$ would remain unreacted in the solution. In this situation, we called $\ce{Na2CO3}$ is the limiting reagent because all of them would be consumed by the reaction. In other word, $\ce{CaCl2}$ would be in excess.

  • $\begingroup$ Okay, your entire solution revolves around a given example, but is missing a conclusion. So, are you implying that whether there would be a limiting reagent or not depends on the amount of the reagents used, correct? $\endgroup$ – Gaurang Tandon May 4 '18 at 12:29
  • $\begingroup$ @ Gaurang Tandon: correct. There is always a limiting reagent, if at least one of the starting materials is in excess in the reaction mixture. $\endgroup$ – Mathew Mahindaratne May 4 '18 at 22:19

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