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SrCl2(aq)+2NaF(aq)⟶SrF2(s)+2NaCl(aq)


What volume of a 0.110 M NaF solution is required to react completely with 167 mL of a 0.420 M SrCl2 solution?


How many moles of SrF2 are formed from this reaction?

Respuesta :

jamesesther1234

Answer:

  • 637.636 mL of NaF
  • 0.7014 moles of SrF2

Explanation:

[tex]C_{1}V_{1}[/tex] = [tex]C_{2}V_{2}[/tex]

Where 1 and 2 represent NaF and SrCl2 respectively; and C and V ae concentration and volume.

0.110 × V1 = 0.420 × 167

V1 = [tex]\frac{70.14}{0.110}[/tex] = 637.636 mL ≈ 0.638 L.

Mole ratio of SrF2:

mole ratio of SrCl2 : SrF2 is 1 : 1 (also, conc × vol = no. of moles )

∴ moles of SrF2 = 0.7014.

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