Question

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?

Answer 1

• 637.636 mL of NaF
• 0.7014 moles of SrF2

Step-by-step explanation:

`C_(1)V_(1)` =
`C_(2)V_(2)`

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

0.110 × V1 = 0.420 × 167

V1 =
`(70.14)/(0.110)` = 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.