Chemistry Home Work Question

Question

A 175.0 mL solution of 2.573 M strontium nitrate is mixed with 200.0 mL of a 3.162 M sodium fluoride solution.



Calculate the mass of the resulting strontium fluoride precipitate.

mass: ______g

Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration.

[Na⁺]= ______M

[NO₃^-]= ______M

[Sr²⁺]= ______M

[F−]= ______M

Answer 1

Sr(NO₃)₂(aq) + 2NaF(aq) ==> SrF₂(s) + 2NaNO₃(aq) ... balanced equation

Find limiting reactant: an easy way is to divide moles of each reactant by the corresponding coefficient in the equation and whichever value is less represents the limiting reactant:

For Sr(NO₃)₂ we have 0.1750 L x 2.573 mol/L = 0.45028 mols (÷1->0.45028)

For NaF we have 0.2000 L x 3.162 mol/L = 0.6324 mols (÷2->0.3162)

Therefore NaF is the limiting reactant

Now, use the limiting reactant to find the moles and mass of SrF₂ produced:

0.6324 mol NaF x 1 mol SrF₂ / 2 mol NaF x 125.6 g SrF₂/mol = 39.71 g SrF₂ produced

The only ion no longer in solution will be the F- ion since it was part of the limiting reactant and also part of the precipitate. To find final concentrations of the other ions, find moles and then divide by the total volume (in liters). The final volume = 0.1750 L + 0.2000 L = 0.3750 L

[Na⁺] = 0.6324 mol NaF x 1 mol Na/mol NaF = 0.6324 mol Na⁺ / 0.3750 L = 1.686 M

[NO₃^-] = 0.4503 mol Sr(NO₃)₂ x 2 mol NO₃^-/mol Sr(NO₃)₂ = 0.9006 mol NO₃^-/0.3750 L = 2.402 M

[Sr²⁺] = 0.4503 mol Sr(NO₃)₂ x 1 mol Sr²⁺/mol Sr(NO₃)₂ = 0.4503 mol Sr²⁺ - 0.3162 mol = 0.1341 mol Sr²⁺/0.3750 L = 0.3576 M

[F-] = 0 M