A 165.0 ml solution of 2.724M strontium nitrate is mixed with 215.0ml of a 3.120M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Question

Assuming the complete precipitation, calculate the final concentration pf each ion. If the ion is no longer in solution, enter a 0 for the concentration.

Answer 1

Sr(NO₃)₂(aq) + 2NaF(aq) ==> SrF₂(s) + 2NaNO₃(aq) .. balanced equation

moles Sr(NO₃)₂ = 165.0 ml x 1 L / 1000 ml x 2.724 mol / L = 0.4495 mols

moles NaF = 215.0 ml x 1 L / 1000 ml x 3.120 mol / L = 0.6708 mols

Because it takes 2 mols NaF to react with 1 mol Sr(NO₃)₂, the NaF is the limiting reactant and will determine the amount of SrF₂ that can be formed.

mols SrF₂ formed = 0.6708 mols NaF x 1 mol SrF₂ / 2 mol NaF = 0.3354 mols SrF₂

molar mass SrF₂ = 125.6 g / mol

mass SrF2 = 0.3354 mols SrF₂ x 125.6 g / mol = 42.13 g SrF₂