Answer:
Mass of the resulting strontium fluoride precipitate = 34.51 g
Step-by-step 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
Answer: 34.51g
Mass of SrF2 = 0.5496/2 moles × 125.6 g/mol = 34.51g SrF2 produced
Step-by-step 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
