Question

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

Answer 1

Answer:

4.07L of a 0.110M NaF are needed

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

Answer 2

Answer:

$V=4.07L$

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:

$n_{SrCl_2}=0.620\frac{mol}{L}*0.361L=0.22molSrCl_2$

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

$n_{NaF}=0.225molSrCl_2*\frac{2molNaF}{1molSrCl_2}=0.448molNaF$

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

$V=\frac{n}{M}=\frac{0.448mol}{0.110mol/L}\\ \\V=4.07L$

Best regards.