Answer to Question #324495 in General Chemistry for Ivan Ackerman

Question

Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown.

SrCl2(aq)+2NaF(aq)⟶SrF2(s)+2NaCl(aq)

What volume of a 0.190 M NaF solution is required to react completely with 417 mL

 of a 0.400 M 0.400 M SrCl2 solution?

How many moles of SrF2 are formed from this reaction?

Answer 1

SrCl_2(aq)+2NaF_(aq)⟶SrF_2(s)+2NaCl_(aq)

C_M = n / V

n (SrCl₂) = 1/2 x n (NaF)

2 x (C_M (NaF) x V (NaF)) = C_M (SrCl₂) x V (SrCl₂)

2 x (0.19 x V (NaF)) = 0.400 x 0.417

V (NaF) = (0.400 x 0.417) / (0.19 x 2) = 0.44 L

n (SrF₂) = n (SrCl₂) = 0.400 x 0.417 = 0.17 mol