Map Sapling Learning macmillan learning A 160.0 mL solution of 2.399 M strontium
ID: 1037243 • Letter: M
Question
Map Sapling Learning macmillan learning A 160.0 mL solution of 2.399 M strontium nitrate is mixed with 200.0 mL of a 2.764 M sodium fluoride solution. Calculate the mass of the resulting strontium fuoride precipitate. Number 34.72 Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration. Number Number Na1.66 Sr.298 Number Number 5.64 x 10-9M J = 112.256 Incorrect. Because NaF is the limiting reactant and the precipitate is SrF2, all fluoride in the original solution will contribute to the precipitate. Therefore the fluoride in the final solution will be essentially zero.Explanation / Answer
Ans. Step 1: Determine Limiting Reactant
# Balanced reaction: Sr(NO3)2(aq) + 2 NaF (aq) -------> SrF2(s) + 2 NaNO3(aq)
Theoretical molar ratio of reactants: Sr(NO3)2 : NaF = 1 : 2
# Moles of Sr(NO3)2 = Molarity x Volume in liters = 2.399 M x 0.160 L = 0.38384 mol
Moles of NaF = 2.764 M x 0.200 L = 0.5528 mol
Experimental molar ratio of reactants: Sr(NO3)2 : NaF = 0.38384 : 0.5528 = 1 : 1.4
Comparing the theoretical and experimental molar ratio of reactant, the experimental moles of NaF is less than its theoretical value of 2 mol while keeping that of Sr(NO3)2 constant at 1 mol.
Hence, NaF is the limiting reactant.
# Step 2: The formation of product follows the stoichiometry of limiting reactant.
Following stoichiometry, 2 mol NaF forms 1 mol SrF2.
So,
Moles of SrF2 formed = (1/2) x Moles of NaF = ½ x 0.5528 mol = 0.2764 mol
Now,
Mass of SrF2 formed = Moles x MW = 0.2764 mol x (125.6168 g/ mol) = 34.72 g
# Step 3: Total volume of mixture = 160.0 mL + 200.0 mL = 360.0 mL
# Being the limiting reactant, all F- ions from NaF is consumed to form SrF2.
The concertation of spectator ions remains unaffected and is given by-
Use C1V1 (original solution) = C2V2 (Mixed solution)
So,
[F-] = 0.0 M
[Na+] = [NaF] = C2 = (2.764 M x 200.0 mL) / 360.0 mL = 1.536 M = 1.54 M
[NO3-] = 2 x [Sr(NO3)2] = (2 x 2.399 M x 160 mL) / 360 mL = 2.13 M
# Step 4: Initial moles of Sr2+ = Moles of Sr(NO3)2 = 0.38384 mol
Moles of Sr2+ consumed = Moles of SrF2 formed = 0.2764 mol
Remaining moles Sr2+ = Initial moles – Moles consumed = 0.38384 mol - 0.2764 mol
= 0.10744 mol
Now,
Remaining [Sr2+] = remaining moles / Vol of mixture in Liters
= 0.10744 mol / 0.360 L
= 0.298 M