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38.8 g is the mass of the resulting strontium fluoride precipitate.
Sr(NO3)2(aq) + 2NaF(aq) ==> SrF2(s) + 2NaNO3 ... (balanced equation)
moles Sr(NO3)2 = 0.180 L x 2.400 mol/L = 0.432 moles
moles NaF = 0.200 L x 3.088 mol/L = 0.6176 moles
NaF is LIMITING REACTANT b/c it takes 2 NaF per 1 Sr(NO3)2 and there isn't enough NaF.
Mass of Sr(NO3)2 precipitate = 0.6176 mol NaF x 1 mol SrF2 / 2 mol NaF x 125.6 g SrF2/mol = 38.8 g
Final volume = 180 ml + 200 ml = 380 ml = 0.380 L
[F-] = 0 since all the F- is precipitated as SrF2
[Sr2+] = 0.432 mol - 0.309 mol = 0.123 mol/0.380 L = 0.324 M
[Na+] = 0.6176 moles / 0.380 L = 1.63 M
[NO3-] = 0.432 moles x 2 = 0.864 mol/0.380 L = 2.27 M
What is a limiting reactant ?
The reactant that is consumed first in a chemical reaction, also known as the limiting reactant (or limiting reagent), restricts the amount of product that may be produced.
The balanced chemical equation's mole ratios are used in each of the several approaches to identify the limiting reactant.
The theoretical yield is the maximum quantity of product that can be produced given the limiting reactant. The actual yield—the quantity of product that is actually gathered—is nearly never greater than the theoretical yield. The percentage yield, which indicates what portion of the theoretical yield was attained, is typically used to convey the actual yield.
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