21. (6 pts) What is the theoretical yield of HF (in g) if 6.00 g CaF2 (FM-78.08)

Question

21. (6 pts) What is the theoretical yield of HF (in g) if 6.00 g CaF2 (FM-78.08) react with 7.20 g HSO (FM = 98.09)? What is the percent yield if 2.63 g of HF (FM = 20.01) are actually obtained? CaF2 + H2SO4 CaSO4 + 2HF Theoretical yield- % yield = 22 (a) (2 pts) What volume (in mL) of 2.50 M LiOH (FM 23.95) is required to make 100.0 mL of 0.670 M LioH? Volume = (b) (1 pt) Write a sentence deseribing how to prepare this solution 23. (6 pts) If 25.00 mL of a vinegar solution (aqueous acetic acid) is neutralized by(i.e, completely reacted with) 23.21 mL of 0.052 M barium hydroxide, what is the molar concentration of acetic acid in the vinegar solution? Molarity

Explanation / Answer

1) CaF2 + H2SO4 --> CaSO4 + 2HF

Molecular weight of CaF2 = 78

Molecular weight of H2SO4 = 98

Molecular weight of CaSO4 = 136

Molecular weight of HF = 20

Here one mole of CaF2 will react with 1 mole of H2SO4 to give 2 moles of HF

so 78 g of CaF2 will react with 98 g of H2SO4 to give 2X20 g of HF

So 6 g of CaF2 will react with 7.5g of H2SO4 to give 3.07 g of HF

here the limiting reagent is H2SO4

so 7.2 g of H2SO4 will react with 5.73 g of CaF2 to give 2.93 g of HF

So theoretical yield = 2.93 g

% yield = actual yield X 100 / theoretical yield = 2.63 X 100 / 2.93 = 89.76 %

22)   Initial concentration of LiOH = M1 = 2.5

Volume of LiOH required = V1

Required concentration of LiOH = M2 = 0.67

Volume of LiOH = V2 = 100

M1V1 = M2V2

2.5 X V1 = 100 X 0.67

V1 = 26.8 mL will be required

b) we can prepare the solution by taking a volumetric flask ( capacity 100mL, will add 26.8 mL of 2.5 M of LiOH and will make up the solution to 100mL .

23)

The volume of vinegar = 25 mL

Volume of Barium hydroxide = 23.21 mL

concentration = 0.052 M = 0.052 X 2 Normal = 0.104 N

applying law of equivalence

Mill equivalent of acid = milli equivalent of base

25 X N1 = 0.104 X 23.21

N1 = normality of acid = molar concentration of acid = 0.0965 Molar

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