Question

Titanium reacts with bromine gas according to the following balanced chemical equation:

2Ti(s)+3Br2(l)→2TiBr3(s)

A: What mass of titanium(III) bromide is produced when 139.29 g titanium reacts with 553.73 g bromine liquid? in gTiBr₃

B: What is the percent yield of titanium(III) bromide if 556.03 g of titanium(III) bromide is actually obtained?

Answer 1

A)

Molar mass of Ti = 47.87 g/mol

mass(Ti)= 139.29 g

use:

number of mol of Ti,

n = mass of Ti/molar mass of Ti

=(1.393*10^2 g)/(47.87 g/mol)

= 2.91 mol

Molar mass of Br2 = 159.8 g/mol

mass(Br2)= 553.73 g

use:

number of mol of Br2,

n = mass of Br2/molar mass of Br2

=(5.5373*10^2 g)/(1.598*10^2 g/mol)

= 3.4651 mol

Balanced chemical equation is:

2 Ti + 3 Br2 ---> 2 TiBr3

2 mol of Ti reacts with 3 mol of Br2

for 2.91 mol of Ti, 4.365 mol of Br2 is required

But we have 3.4651 mol of Br2

so, Br2 is limiting reagent

we will use Br2 in further calculation

Molar mass of TiBr3,

MM = 1*MM(Ti) + 3*MM(Br)

= 1*47.87 + 3*79.9

= 287.57 g/mol

According to balanced equation

mol of TiBr3 formed = (2/3)* moles of Br2

= (2/3)*3.4651

= 2.3101 mol

use:

mass of TiBr3 = number of mol * molar mass

= 2.3101* 287.57

= 664.31 g

Answer: 664.31 g

B)

% yield = actual mass*100/theoretical mass

= 556.03*100/664.31

= 83.700 %

Answer: 83.700 %