Question

3. Fluorine reacts with vanadium to produce vanadium(V) fluoride according to the following equation: 5 F2 (g) + 2 V (s) → 2 VF5 (l) (a) How many moles of vanadium are needed to react completely with 0.1917 g of fluorine? (b) What is the maximum theoretical mass of vanadium(V) fluoride that can be produced? (c) If 0.225 g of vanadium(V) fluoride is obtained, what is the percent yield of VF5 (l)? Molar masses (g/mol): F2 38.00 VF5 145.93

Answer 1

5 F₂ (g) + 2 V (s) → 2 VF₅ (l)

Here

2 moles of vanadium(V) reacts with 5 moles of fluorine (F₂)

(2/5) moles of vanadium(V) reacts with 1 moles of fluorine (F₂)

Molar mass of fluorine (F₂) = 38 g/mole

So, 0.1917 g of fluorine = (0.1917 g) / (38 g/mole) = 5.04 x 10^-3 mole

1 moles of fluorine (F₂) reacts with (2/5) moles of vanadium(V).

5.04 x 10^-3 moles of fluorine (F₂) reacts with [(2/5) x 5.04 x 10^-3] moles of vanadium(V).

5.04 x 10^-3 moles of fluorine (F₂) reacts with 2.02 x 10^-3 moles of vanadium(V).

(b)

2 moles of vanadium(V) produces 2 moles of vanadium(V) fluoride (VF₅)

So, 2.02 x 10^-3 moles of vanadium(V) will produces 2.02 x 10^-3 moles of vanadium(V) fluoride (VF₅)

Molar mass of VF₅ = 145.93 g/mole

So, 2.02 x 10^-3 moles of vanadium(V) will produces 0.3 gm of vanadium(V) fluoride (VF₅).

(c)

Compound obtained =0.225 g

Theoretical yield = 0.3

So, % yield = (Compound obtained/ Theoritical yield) x 100

                   = (0.225/0.3) x 100 = 75 %