# 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!

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