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

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.