Question

A 150.0 mL solution of 2.671 M strontium nitrate is mixed with 210.0 mL of a 2.617 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Answer 1

Answer: 34.51g

Mass of SrF2 = 0.5496/2 moles × 125.6 g/mol = 34.51g SrF2 produced

Explanation:

Firstly we need to write a balanced chemical equation for the reaction:

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

We need to determine the limiting reactant:

moles Sr(NO3)2 present = 0.150 L x 2.671 mol/L = 0.4007 moles

moles NaF = 0.210 L x 2.617 mol/L = 0.5496 moles

To determine the limiting reactant:

Limiting reactant = NaF because we need two times the moles of NaF to react with Sr(NO3)2

We don't have enough NaF to react with Sr(NO3)2

(Notice that from the chemical equation the mole ratio is 1:2)

Mass SrF2 produced:

2 moles of NaF will produce 1 mole of SrF2(s)

0.5496 mol NaF will produce mol 0.5496/2 moles of SrF2

Mass of SrF2 = 0.5496/2 moles × 125.6 g/mol = 34.51g SrF2 produced

Answer 2

Mass of the resulting strontium fluoride precipitate = 34.51 g

Explanation:

First, write a correctly balanced equation for the reaction taking place:

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

Next, Find the limiting reactant:

moles Sr(NO3)2 present = 0.150 L x 2.671 mol/L = 0.40065 moles

moles NaF = 0.210 L x 2.617 mol/L = 0.54957 moles

LIMITING REACTANT = NaF

because you need twice the moles of NaF as you have Sr(NO3)2 (see bal. eq.)

Mass SrF2 producedb =

0.54957 mol NaF x 1 mol SrF2 / 2 mol NaF x 125.6 g/mol = 34.51 g SrF2 produced