Final answer:
The mass of the strontium fluoride precipitate formed when a solution of strontium nitrate is mixed with a solution of sodium fluoride is 49.74 grams. Strontium nitrate is the limiting reactant in this double replacement reaction.
Step-by-step explanation:
To calculate the mass of the resulting strontium fluoride precipitate when a solution of strontium nitrate is mixed with a solution of sodium fluoride, we first need to determine which reagent is the limiting reactant. The reaction between strontium nitrate (Sr(NO3)2) and sodium fluoride (NaF) is a double replacement reaction that forms strontium fluoride (SrF2) and sodium nitrate (NaNO3):
Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq)
The reaction stoichiometry shows that one mole of strontium nitrate reacts with two moles of sodium fluoride to form one mole of strontium fluoride. Now, we'll calculate the number of moles of each reactant:
- Strontium nitrate: 185.0 mL × 2.139 mol/L = 0.3957 moles
- Sodium fluoride: 220.0 mL × 2.420 mol/L = 0.5324 moles
According to the stoichiometry, we require twice as many moles of NaF as Sr(NO3)2 to react completely. Hence, strontium nitrate is the limiting reactant.
Next, we calculate the mass of SrF2 precipitate:
Molar mass of SrF2 = 87.62 g/mol (Sr) + 2 × 19.00 g/mol (F)
= 125.62 g/mol
Mass of SrF2 = 0.3957 moles × 125.62 g/mol
= 49.74 g
The mass of the strontium fluoride precipitate is 49.74 grams.