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 0.110 M NaF solution is required to react completely with 361 mL of a 0.620 M SrCl2 solution?

Answer 1

`V=4.07L`

Step-by-step explanation:

Hello,

In this case, for the proposed question, we should notice that the moles of sodium fluoride and strontium chloride, must be in stoichiometric proportions, that is 1:2 respectively, based on the given chemical reaction, for that reason, we first compute the available moles of strontium chloride based on the statement:

`n_(SrCl_2)=0.620(mol)/(L)*0.361L=0.22molSrCl_2`

Now, based on the reaction, the moles of sodium fluoride which will complete react result:

`n_(NaF)=0.225molSrCl_2*(2molNaF)/(1molSrCl_2)=0.448molNaF`

Finally, by using the molarity of the sodium fluoride solution, we compute the required volume of such solution:

`V=(n)/(M)=(0.448mol)/(0.110mol/L)\\ \\V=4.07L`

Best regards.

Answer 2

4.07L of a 0.110M NaF are needed

Step-by-step explanation:

Based on the reaction:

SrCl₂(aq)+2NaF(aq)⟶SrF₂(s)+2NaCl(aq)

1 mole of strontium chloride react with 2 moles of NaF

361mL of 0.620M SrCl₂ solution has:

0.361L ₓ (0.620mol / L) = 0.22382 moles SrCl₂.

Moles of NaF for a complete reaction must be:

0.22382 moles SrCl₂ ₓ (2 mol NaF / 1 mol SrCl₂) = 0.44764 moles of NaF

If you have a solution of 0.110M NaF, the moles of NaF needed are:

0.44764 moles of NaF ₓ (1L / 0.110mol NaF) = 4.07L of a 0.110M NaF are needed