I clearly need a lot of help with this one. I would be extremely grateful if you

Question

I clearly need a lot of help with this one. I would be extremely grateful if you could help me by getting the right answer and possibly giving a short explanation on how you got the answer. Thanks in advance!

Elemental fluorine, F2, can oxidize just about any metal, including, as seen below, titanium (Ti): 1S points 2F,(g) + Ti(s) > TiF4 (s) (1) Given that one begins with 3.19 L of fluorine (at 15.00 atm and 298K) and 52.4 g of ttanium etermine the number of equivalents of each reactant and indicate which species is the limiting reactant MO MD 47.D 41.85-(n) 24.4 2 Mol 2-2. Eq(Ti) Limiting reactant: (2) Assuming the same starting amounts as in part (1), what is the actual yield of TiF(s) in grams if the percent yield of the process is only 87%? 817 41AD Ti Mol l 121-125 02 MD MO Actual yield: 252 TiF4

Explanation / Answer

1. Using n = PV/RT

with,

P = 15 atm

V = 3.19 L

T = 298 K

R = gas constant

the number of moles of F2 = 15 x 3.19/0.08206 x 298 = 1.96 mols

moles of Ti = 52.4 g/47.87 g/mol = 1.09 mols

Number of equivalents of (F2 : Ti) = (0.64 : 0.36) = (1.778 : 1.00)

If all of Ti is consumed we would need = 2 x 1.09 = 2.18 mols of F2

If all of F2 is consumed we would need = 1.96/2 = 0.98 mols of Ti

Since moles of F2 is less than required, this is the limiting reactant in the reaction.

(2) Theoretical yield of TiF4 = (1.96 mol/2) x 123.861 g/mol = 121.384 g (100%)

Percent yield found 87% = 0.87 x 121.384 = 105.60 g of TiF4 formed

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