# How to calculate the amount of substance of barium sulfate that is precipitated?

Determine the amount (in mol) of barium sulfate that will be precipitated when $200.0~\mathrm{cm^3}$ of $0.450~\mathrm{mol\,dm^{–3}}$ barium nitrate solution is added to an excess of sodium sulfate solution, given that the equation for the reaction is: $$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)}$$

I have just tried doing it and I got an answer of $0.09~\mathrm{mol}$, but I am not sure if this is correct, because I have only just been introduced to $\mathrm{dm^{-3}}$ so didn't really know what I was doing. I converted $200~\mathrm{cm^3}$ to $0.2~\mathrm{dm^3}$ (is this right?), and then multiplied this value by the concentration, $0.450~\mathrm{mol\,dm^{-3}}$, and this gave me $0.09~\mathrm{mol}$. Since the stoichiometric ratio between $\ce{BaNO3}$ and $\ce{BaSO4}$ is 1:1, I just suspected that the amount of substance of $\ce{BaSO4}$ would also be $0.09~\mathrm{mol}$.

If anyone could tell me if I have done this correctly, it would be greatly appreciated.

• Yep! That's basically correct :) – IT Tsoi May 18 '16 at 13:40

## 1 Answer

Yes, your answer is correct.

However, since you are still learning I recommend, that you take the time to write out all the equations, with proper units and every step. This should help you to gain a better understanding of the chemistry that is actually going on.

One of the key words of the exercise is excess of sodium sulfate solution, which indicates that the reaction will be complete, i.e. entirely on the product side.

Hence, you can (and you correctly did) determine from the reaction equation, $$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)},$$ that the stoichiometric ratio is one, $n(\ce{Ba(NO3)2})/n(\ce{BaSO4}) = 1$. Therefore you can also write $$n(\ce{Ba(NO3)2}) = n(\ce{BaSO4}).$$

Now you simply have to calculate the amount of substance of barium nitrate. You have been given the volume of your solution $V(\ce{Ba(NO3)2}) = 0.200~\mathrm{cm^3}$ and the concentration $c(\ce{Ba(NO3)2}) = 0.450~\mathrm{mol\,dm^{-3}}$. Using the relation $$c = \frac{n}{V}$$ you can calculate the amount of substance of barium nitrate.

You have also correctly converted cubic centimetres to cubic decimetres. In any case, you should get familiar with metric prefixes since they will come up in every science class you will take.