Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown. srcl2(am)+2naf(aq) —-> srf2(s)+2nacl(aq). what volume of a 0.510 m naf solution is required to react completely with 439 ml of a 0.660 m srcl2 solution

1,14L of 0.510M NaF solution

Explanation:

For the reaction:

SrCl₂(aq) + 2NaF(aq) → SrF₂(s) + 2NaCl(aq)

Moles in 439mL, 0,439L of a 0.660M SrCl₂ are:

0,439L×0,660mol/L = 0,290 moles of SrCl₂

As 1 mole of SrCl₂ reacts with 2 moles of NaF:

0,290 moles of SrCl₂×(2mol NaF / 1mol SrCl₂) = 0,580mol NaF.

These moles of NaF are required for a complete reaction with SrCl₂. In a 0.510M NaF solution you have 0.580mol in:

0,580mol × (1L/0,510mol) = 1,14L of 0.510M NaF solution

I hope it helps!