Chemistry - A 170.0 mL solution of 2.773 M strontium nitrate is mixed with 200.0 mL of a 3.354 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Question

help me please, A 170.0 mL solution of 2.773 M strontium nitrate is mixed with 200.0 mL of a 3.354 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Answer 1

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

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

Next, Find the limiting reactant:

moles Sr(NO₃)₂ present = 0.170 L x 2.773 mol/L = 0.4714 moles

moles NaF = 0.200 L x 3.354 mol/L = 0.6708 moles

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

Mass SrF₂ produced: 0.6708 mol NaF x 1 mol SrF2/2 mol NaF x 125.6 g/mol = 42.13 g SrF₂ produced