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

Answer –

We are given reaction –

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

We need to calculate theoretical yield from the given initial amount of reactants.

Part A) 5.0 g Ti and 5.0 g F2

First we need to calculate moles of both reactants –

Moles of Ti = 5.0 g / 47.867 g.mol-1

                    = 0.104 moles

Moles of F2 = 5.0 g / 37.996 g.mol-1

                    = 0.132 mole

Now we need to calculate limiting reactant

Moles of TiF4 from the Ti

From the balanced equation –

1 moles of Ti = 1 moles of TiF4

So, 0.104 moles of Ti = ?

= 0.104 moles of TiF4

Moles of TiF4 from F2

From the balanced equation

2 moles of F2 = 1 moles of TiF4

So, 0.132 moles of F2 = ?

= 0.0658 moles of TiF4

So moles of TiF4 is lowest from the F2, so F2 is the limiting reactant

So, moles of TiF4 = 0.0658 moles

Theoretical yield of TiF4 = 0.0658 mole * 123.859 g/mole

                                         = 8.1 g

Part B) 2.2 g Ti and 1.7 g F2

First we need to calculate moles of both reactants –

Moles of Ti = 2.2 g / 47.867 g.mol-1

                    = 0.0459 moles

Moles of F2 = 1.7 g / 37.996 g.mol-1

                    = 0.0447 mole

Now we need to calculate limiting reactant

Moles of TiF4 from the Ti

From the balanced equation –

1 moles of Ti = 1 moles of TiF4

So, 0.0459 moles of Ti = ?

= 0.0459 moles of TiF4

Moles of TiF4 from F2

From the balanced equation

2 moles of F2 = 1 moles of TiF4

So, 0.0447 moles of F2 = ?

= 0.0224 moles of TiF4

So moles of TiF4 is lowest from the F2, so F2 is the limiting reactant

So, moles of TiF4 = 0.0224 moles

Theoretical yield of TiF4 = 0.0224 mole * 123.859 g/mole

                                         = 2.8 g

Part C) 0.227 g Ti and 0.292 g F2

First we need to calculate moles of both reactants –

Moles of Ti = 0.227 g / 47.867 g.mol-1

                    = 0.00474 moles

Moles of F2 = 0.292 g / 37.996 g.mol-1

                  = 0.00768 mole

Now we need to calculate limiting reactant

Moles of TiF4 from the Ti

From the balanced equation –

1 moles of Ti = 1 moles of TiF4

So, 0.0074 moles of Ti = ?

= 0.00474 moles of TiF4

Moles of TiF4 from F2

From the balanced equation

2 moles of F2 = 1 moles of TiF4

So, 0.00768 moles of F2 = ?

= 0.00384 moles of TiF4

So moles of TiF4 is lowest from the F2, so F2 is the limiting reactant

So, moles of TiF4 = 0.00384 moles

Theoretical yield of TiF4 = 0.00384 mole * 123.859 g/mole

                                         = 0.476 g

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