Precipitation reaction stoichiometry?
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.