Question

For the following reaction, 30.1 grams of bromine are allowed to react with 11.3 grams of chlorine gas bromine(g) + chlorine(g) bromine monochloride(g) What is the maximum mass of bromine monochloride that can be formed?grams What is the FORMULA for the limiting reagent? whit mass ofthe excessreagent remains after the reaction is complete?[ コgrams Suinit Ansur roupmore group attmgt Retry Entire 9 more group attempts remaining

Answer 1

1)

Molar mass of Br2 = 159.8 g/mol

mass(Br2)= 30.1 g

use:

number of mol of Br2,

n = mass of Br2/molar mass of Br2

=(30.1 g)/(1.598*10^2 g/mol)

= 0.1884 mol

Molar mass of Cl2 = 70.9 g/mol

mass(Cl2)= 11.3 g

use:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(11.3 g)/(70.9 g/mol)

= 0.1594 mol

Balanced chemical equation is:

Br2 + Cl2 ---> 2 BrCl +

1 mol of Br2 reacts with 1 mol of Cl2

for 0.1884 mol of Br2, 0.1884 mol of Cl2 is required

But we have 0.1594 mol of Cl2

so, Cl2 is limiting reagent

we will use Cl2 in further calculation

Molar mass of BrCl,

MM = 1*MM(Br) + 1*MM(Cl)

= 1*79.9 + 1*35.45

= 115.35 g/mol

According to balanced equation

mol of BrCl formed = (2/1)* moles of Cl2

= (2/1)*0.1594

= 0.3188 mol

use:

mass of BrCl = number of mol * molar mass

= 0.3188*1.154*10^2

= 36.77 g

Answer: 36.8 g

2)

Cl2 is limiting reagent

3)

According to balanced equation

mol of Br2 reacted = (1/1)* moles of Cl2

= (1/1)*0.1594

= 0.1594 mol

mol of Br2 remaining = mol initially present - mol reacted

mol of Br2 remaining = 0.1884 - 0.1594

mol of Br2 remaining = 2.898*10^-2 mol

Molar mass of Br2 = 159.8 g/mol

use:

mass of Br2,

m = number of mol * molar mass

= 2.898*10^-2 mol * 1.598*10^2 g/mol

= 4.631 g

Answer: 4.63 g