# Precipitation reaction stoichiometry?

I have a question where I do not understand the final step to solve...here it is:

What will be the Strontium ion concentration remaining after 30.0 mL of 0.10 M Na2SO4 solution are added to 70.0 mL of 0.20 M Sr(NO3)2 solution?

First, I wrote the equation: Sr + SO4 -> SrSo4 (I know this is excluding ions, because I am primarily focused on the stoichiometry part)

I figured moles of Na2So4 and Sr(NO3)2, which correspond to moles of strontium. These are .003 moles Na2SO4 and 0.014 moles Sr(NO3)2. This is where I got stuck, looked to my book's answer explanation, and got more confused. The book states "0.0030 mol of sulfate ion will combine with 0.0030 mol of strontium ion, leaving 0.011 mol of strontium in a total volume of 100.0 mL."

Why is the reaction leaving 0.011 moles of strontium? What was the calculation that obtained this number?

Thank you!

## 1 Answer

Sr(NO3)2 + Na2SO4 = SrSO4 + 2NaNO3

So, here we have, 1 mol Sr(NO3)2 reacts with 1 mol of Na2SO4 by unitary method, you can say, 0.003 mol Sr(NO3)2 reacts with 0.003 mol of Na2SO4. But checking the reactants we have 0.003 moles Na2SO4 and 0.014 moles Sr(NO3)2

So only 0.003 moles of each reacts with the other to leave out any excess in either of the reactants, as in this case, Sr(NO3)2.

When 0.003 mol reacts of Sr(NO3)2, we have $$0.014 (initial) - 0.003 (used) = 0.011mol$$ of Sr(NO3)2 left.

• If you liked the answer, you could also upvote it. :) @tyger2020 – CodeBlooded Mar 2 '18 at 14:16