I have the correct answer (1.199; 1.094; Ti; 117.9g) however I am having issues

Question

I have the correct answer (1.199; 1.094; Ti; 117.9g) however I am having issues figuring out how the answer was found (my answers when I try to attempt to solve the problem are way off). Can anyone please help me with the process? If possible, please show work/formulas

Elemental fluorine, F2, can oxidize just about any metal, including, as seen below, titanium (Ti): 15 points 2F2(g)+ Ti(s) TiF (s) (1) Given that one begins with 3.91 L of fluorine (at 15.00 atm and 298K) and 52.4 g of titanium, determine the number of equivalents of each reactant and indicate which species is the limiting reactant Eq(Ti: Limiting reactant: (2) Assuming the same starting amounts as in part (1), what is the actual yield of TiF

Explanation / Answer

1)
For TI:
Molar mass of Ti = 47.867 gm
mass of Ti = 52.4 gm
Number of moles of Ti = 52.4/47.867 =1.095

FOR F2:
use:
P*V=n*R*T
15*3.91= n*(0.0821)*(298)
n=2.4 mol

2 mol of F2 requires 1 mol of Ti
2.4mol of F2 will need 2.4/2 = 1.2 molTi
But Ti is only 1.095 mol

Limiting reagent: Ti
Eq(Ti)= 1.095
Eq(F2) = 2.4

2.
1mol of Ti forms 1 mol of TiF4
So number of mols of TIF4 formed = 1.095 mol
Molar mass of TiF4 = 123.861 gm
mass of TiF4 formed theoretically = number of moles * molar mass
   = 1.095*123.861
   = 135.63 gm

But actual yield is 87 %
Actual yield= 87% of 135.63
    = 117.9 gm

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