Question

For the reaction Ti(s)+2F2(g)→TiF4(s) compute the theoretical yield of the product (in grams) for each of the following initial amounts of reactants. Part A 2.5 g Ti, 1.8 g F2 Express your answer using two significant figures. Part B 0.235 g Ti, 0.286 g F2 Express your answer using two significant figures.

Answer 1

A)

Molar mass of Ti = 47.87 g/mol

mass(Ti)= 2.5 g

use:

number of mol of Ti,

n = mass of Ti/molar mass of Ti

=(2.5 g)/(47.87 g/mol)

= 5.222*10^-2 mol

Molar mass of F2 = 38 g/mol

mass(F2)= 1.8 g

use:

number of mol of F2,

n = mass of F2/molar mass of F2

=(1.8 g)/(38 g/mol)

= 4.737*10^-2 mol

Balanced chemical equation is:

Ti + 2 F2 ---> TiF4

1 mol of Ti reacts with 2 mol of F2

for 5.222*10^-2 mol of Ti, 0.1044 mol of F2 is required

But we have 4.737*10^-2 mol of F2

so, F2 is limiting reagent

we will use F2 in further calculation

Molar mass of TiF4,

MM = 1*MM(Ti) + 4*MM(F)

= 1*47.87 + 4*19.0

= 123.87 g/mol

According to balanced equation

mol of TiF4 formed = (1/2)* moles of F2

= (1/2)*4.737*10^-2

= 2.368*10^-2 mol

use:

mass of TiF4 = number of mol * molar mass

= 2.368*10^-2*1.239*10^2

= 2.934 g

Answer: 2.9 g

B)

Molar mass of Ti = 47.87 g/mol

mass(Ti)= 0.235 g

use:

number of mol of Ti,

n = mass of Ti/molar mass of Ti

=(0.235 g)/(47.87 g/mol)

= 4.909*10^-3 mol

Molar mass of F2 = 38 g/mol

mass(F2)= 0.286 g

use:

number of mol of F2,

n = mass of F2/molar mass of F2

=(0.286 g)/(38 g/mol)

= 7.526*10^-3 mol

Balanced chemical equation is:

Ti + 2 F2 ---> TiF4

1 mol of Ti reacts with 2 mol of F2

for 4.909*10^-3 mol of Ti, 9.818*10^-3 mol of F2 is required

But we have 7.526*10^-3 mol of F2

so, F2 is limiting reagent

we will use F2 in further calculation

Molar mass of TiF4,

MM = 1*MM(Ti) + 4*MM(F)

= 1*47.87 + 4*19.0

= 123.87 g/mol

According to balanced equation

mol of TiF4 formed = (1/2)* moles of F2

= (1/2)*7.526*10^-3

= 3.763*10^-3 mol

use:

mass of TiF4 = number of mol * molar mass

= 3.763*10^-3*1.239*10^2

= 0.4661 g

Answer: 0.47 g