Using the balanced reaction equation, we can calculate that the mass of the precipitate will be 1.617 g.
To solve this question, we need to write the balanced reaction equation first:
2 NaF(aq) + BeSO₄(aq) → Na₂SO₄(aq) + BeF₂(s)
We can now see that the precipitate will be beryllium fluoride. We can also see that 1 mol of beryllium sulfate reacts with 2 moles of sodium fluoride. Now we can calculate the number of moles (n) of both reactants using their molarities (c) and volumes (V):
c = n/V ⇒ n = c * V
n(NaF) = c(NaF) * V(NaF)
n(NaF) = 4.988 M * 79.854 mL = 0.3983 mol
n(BeSO₄) = c(BeSO₄) * V(BeSO₄)
n(BeSO₄) = 1.471 M * 23.3760 mL
n(BeSO₄) = 0.03439 mol
We can see that n(NaF) is more than 2 times greater than n(BeSO₄), so NaF is in excess, and BeSO₄ is the limiting reactant. Now we can calculate the number of moles of BeF₂ formed:
1 mol BeSO₄ : 1 mol BeF₂ = 0.03439 mol BeSO₄ : X
X = 0.03439 mol BeF₂
We can now convert the number of moles (n = 0.03439 mol) of BeF₂ into mass (m), using the molar mass of BeF₂ (M = 47.01 g/mol):
n = m/M ⇒ m = n * M
m = 0.03439 mol * 47.01 g/mol
m = 1.617 g
You can learn more about balanced reaction equations here:
brainly.com/question/22064431
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