Question

Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown. SrF2 (s)2 NaCl(aq) SrCl2(aq)2 NaF(aq) What volume of a 0.610 M NaF solution is required to react completely with 945 mL of a 0.700 M SrCl, solution? volume: mL How many moles of SrF, are formed from this reaction? moles of SrF2 mol

Answer 1

1)

Here:

M(SrCl2)=0.7 M

M(NaF)=0.61 M

V(SrCl2)=945.0 mL

According to balanced reaction:

2*number of mol of SrCl2 =1*number of mol of NaF

2*M(SrCl2)*V(SrCl2) =1*M(NaF)*V(NaF)

2*0.7 M *945.0 mL = 1*0.61M *V(NaF)

V(NaF) = 2169 mL

Answer: 2.17*10^3 mL

2)

lets calculate the mol of SrCl2

volume , V = 9.45*10^2 mL

= 0.945 L

use:

number of mol,

n = Molarity * Volume

= 0.7*0.945

= 0.6615 mol

According to balanced equation

mol of SrF2 formed = (2/1)* moles of SrCl2

= (2/1)*0.6615

= 1.323 mol

Answer: 1.32 mol