Question

Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown: SrCl₂(aq) + 2NaF(aq) → SrF2(s) + 2NaCl(aq). What volume of a 0.250 M NaF solution is required to react completely with 445 mL of a 0.580 M SrCl₂ solution? How many moles of SrF2 are formed from this reaction?

Answer 1

To react completely with 445 mL of 0.580 M SrCl₂, you need 2.0648 liters of 0.250 M NaF. The amount of SrF₂ formed is 0.2581 mol.

Step-by-step explanation:

The question is about finding the volume of a 0.250 M NaF solution that is required to react completely with 445 mL of a 0.580 M SrCl₂ solution, and the amount of SrF₂ that is formed as a result of the reaction. First, we need to determine the moles of SrCl₂ used in the reaction using the concentration and volume of SrCl₂. The reaction shows that one mole of SrCl₂ reacts with two moles of NaF, so we can use stoichiometry to find the required amount of NaF.

We calculate: Moles of SrCl₂ = 0.580 mol/L × 0.445 L = 0.2581 mol. Since 1 mole of SrCl₂ reacts with 2 moles of NaF, we need 2 × 0.2581 mol = 0.5162 mol of NaF. Now we can find the volume of 0.250 M NaF solution needed: Volume of NaF = Moles of NaF / NaF Molarity = 0.5162 mol / 0.250 M = 2.0648 L. This means that 2.0648 liters (or 2064.8 mL) of 0.250 M NaF solution are required to completely react with 445 mL of 0.580 M SrCl₂ solution.

The amount of SrF₂ formed is equal to the moles of SrCl₂, which is 0.2581 mol, assuming it goes to completion due to the stoichiometry of the reaction.