For a lab assignment, I combined $\pu{10 mL}$ of $\pu{1 M}$ sodium sulfate solution with $\pu{10 mL}$ of $\pu{1 M}$ calcium chloride solution. The ionic equation I came up with is:

$$\begin{multline} \ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + 2Cl^- (aq) ->}\\ \ce{CaSO4(s) + 2Na^+ (aq) + 2Cl^- (aq)} \end{multline}$$

I was then asked, in a lab question, the following:

If the reaction called for $\pu{1 M}$ chloride ion with $\pu{1 M}$ calcium ion, would you still use $\pu{10 mL}$ of each solution for a complete reaction? Explain your reasoning.

To my understanding, I initially had $\pu{2 M}$ chloride ion and $\pu{1 M}$ calcium ion. With only $\pu{1 M}$ chloride ion instead, the ionic equation would change to this, or so I'm led to believe:

$$\begin{multline} \ce{2Na^+ (aq) + SO4^2- (aq) + Ca^2+ (aq) + Cl^- (aq) ->}\\ \ce{CaSO4(s) + 2Na^+ (aq) + Cl^- (aq)} \end{multline}$$

As a (balanced) molecular equation, to my understanding, this would be:

$$\begin{multline} \ce{Na2SO4 (aq) + Ca (aq) + Cl (aq) ->}\\ \ce{CaSO4(s) + NaCl (aq) + Na (aq)} \end{multline}$$

Is this a complete reaction? I have seen conflicting definitions of a complete reaction online:

  • answers.com says that a complete reaction means all of the reactants are reacted into products.
  • However, someone on chemicalforums.com says that a complete reaction means that all of the limiting reactant is used up. The University School of Milwaukee seems to back up what the user on chemicalforums.com said ("all of at least one of the available reactants is used up and converted into products").

My main question is thus: what is a complete reaction? No matter my final equation for this lab question, I still need to know what a complete reaction is so that I can identify them in general.

My intuition says that this is a complete reaction, since all of the calcium ions and the chloride ions are converted to products. Is this correct? The leftover sodium ion makes me suspicious, so I don't know how to answer.

I later spoke to my lab instructor and he clarified that the question is asking for me to come up with two arbitrary aqueous solutions, one which contains $\pu{1M}$ of chloride ion and another which contains $\pu{1M}$ of calcium ion, e.g. $\ce{KCl(aq)}$ and $\ce{Ca(NO3)2 (aq)}$.

So in this case, I would need twice the volume of potassium chloride to get the number of chloride ions I need to balance the equation and prevent explosions, i.e. no extra sodium ions leftover.

  • $\begingroup$ This question has an archived discussion in chat. $\endgroup$
    – 2rs2ts
    Commented Mar 19, 2014 at 14:22

1 Answer 1


I suspect that you would have both these molecular equations:

\begin{align} \ce{2KCl(aq) + Na2SO4(aq) &-> K2SO4(aq) + 2NaCl(aq)}\\ \ce{Ca(NO3)2(aq) + Na2SO4(aq) &-> 2NaNO3(aq) + CaSO4(aq)} \end{align}

To answer your question with "What is a complete reaction?":

A complete reaction is one where all of at least one of the available reactants is used up and converted into products.

So for example $\ce{KCl}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{2 M}$ $\ce{KCl}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. However, you only have $\pu{1 M}$ $\ce{KCl}$ (so $\pu{1 M}$ chloride ions). This means that, as you mentioned, you will need the double concentration for the $\ce{KCl}$ solution to achieve a complete reaction.

So in this case the only way for all $\ce{Na2SO4}$ molecules to react is to add the double amount of $\ce{KCl}$ solution. Then this is a complete reaction : at least one reactant is used up in the reaction.

But say we have a situation in which we have a $\pu{2 M}$ $\ce{KCl}$ solution, but a $\pu{2 M}$ $\ce{Na2SO4}$ solution, then all the $\ce{KCl}$ would react, but still \pu{1 M} of $\ce{Na2SO4}$ will not have reacted.If in your lab assignment it is the meaning that you react all $\ce{Na2SO4}$, then in this case you don't have a complete reaction. Only when a two times larger volume of the $\ce{KCl}$ is added (to 2M $\ce{Na2SO4}$), then you will once again achieve a complete reaction. Furthermore you would just get a solution of solvated ions.

In the case of $\ce{Ca(NO3)2}$ and $\ce{Na2SO4}$: In this equation you will need $\pu{1 M}$ $\ce{Ca(NO3)2}$ and $\pu{1 M}$ $\ce{Na2SO4}$ in order for all the reactants to react. You have $\pu{10 ml}$ of both compounds, each $\pu{1 M}$ in concentration: so in this case there is no need to use a larger volume of aqueous $\ce{Ca(NO3)2}$.


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