Exercise 4.46 For the reaction Ti(s) 2 F 2 (g) TiF 4(s) compute the theoretical

Question

Exercise 4.46 For the reaction Ti(s) 2 F 2 (g) TiF 4(s) compute the theoretical yield of the product (in grams) for each of the following initial amounts of reactants. You may want to reference (DA ages 145-152) Section 4.3 while completing this problem. Part A 5.0 g Ti, 5.0 g F2 Express your answer using two significant figures. Submit My Answers Give U Part B 2.2 g Ti, 1.7 g F2 Express your answer using two significant figures. TTE Submit My Answers Give U Part C 0.227 g Ti, 0.292 g F2 Express the mass in grams to three significant figures. Submit My Answers Give U

Explanation / Answer

Theoretical yield is the maximum amount of product that can be formed from the amount of reagent given.

For each one you need to work out the number of moles of each reagent

From this you can work out the limiting reagent, which is the reagent that is not present in enough quantity to fully react will all of the other reagent

the moles of product formed will the be the amount that forms if all of the limiting reagent is converted to product

Then convert moles of product to mass

1.
Ti(s) + 2F2(g) ----> TiF4

moles = mass / molar mass
molar mass Ti = 47.90 g/mol
molar mass F2 = (2 x 19.0) = 38.0 g/mol
molar mass TiF4 = 47.90 + (4 x 19.0) = 123.9 g/mol

moles Ti = mass / molar mass = 5.0 g / 47.90 g/mol
= 0.104 mole Ti

moles F2 = 5.0 g / 38.0 g/mol
= 0.131 mol F2

Now, equation shows that
1 mol of Ti requires 2 moles of F2 to fully react
Therefore 0.146 mol Ti will require (2 x 0.104) mol of F2 in order to fully react = 0.208 mol of F2

So the amount of F2 required to react all the Ti given is 0.208 moles

but we only have 0.131 mol of F2, which is not enough to fully react with the Ti

That means the F2 is the limiting reagent and Ti is in excess.

The max amount of product possible is if all the limiting reagent is converted to product.

Equation states that
2 moles of F2 --------> 1 mole of TiF4
therefore
1 mole F2 -----> 1/2 moles TiF4
thus 0.131 mole F2 ----> (1/2 x 0.131) mol TiF4
= 0.65 moles TiF4 can be formed from 0.131 mol (5.0g) F2

mass of TiF4 = moles x molar mass
= 0.065 mol x 123.9 g/mol
= 8.05 g (2 sig fig)

Similarly, for the others.

Part B: 2.77 g

Part C: 0.291 g

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