Question

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

Answer 1

The required volume of NaF is 765.97 mL.

0.18 moles of SrF2 are formed.

Volume calculation

- First, we need to know the molar mass of the reactants:

SrCl2 molar mass = 158.53 g/mol

NaF molar mass = 41.98 g/mol

- Second, we need to know the amount of moles that are in the 573mL of SrCl2:

`\frac{573mL\text{ . 0.320 moles}}{1000mL}=0.18\text{moles}`

So, there are 0.18 moles of SrCl2 in the 573mL of a 0.320 M solution of SrCl2.

- Now, we relate the stoichiometry of the equation to the number of moles of SrCl2 that we have:

`\frac{0.18\text{ moles . 2 moles}}{1\text{ mol}}=\text{ 0.36 moles}`

So, we need 0.36 moles of NaF.

- Finally, we calculate the volume of 0.470M NaF that is needed to react completely with the SrCl2:

`\frac{0.36\text{ moles . 1000mL}}{0.470\text{ moles}}=\text{ 765.97 mL}`

So, the required volume of NaF is 765.97 mL.

Moles of SrF2 formed

From the balanced equation, we know that 1 mole of SrF2 is formed from 1 mole of SrCl2, so we calculate the moles of SrF2 formes from the 0.18 moles of the SrCl2 solution:

`\frac{0.18\text{ moles . 1 mol}}{1mol_{}}\text{ = 0.18 moles}`

As the relation is 1 to 1, 0.18 moles of SrF2 are formed.