Chloroform liquid, CHCI3, is formed by reaction of methane gas, CH4, and chlorin

Question

Chloroform liquid, CHCI3, is formed by reaction of methane gas, CH4, and chlorine gas, CI2. The reaction also produces gaseous hydrochloric acid, HCI. The molar mass of the species are as follows, CHCI3 119.4 g/mol; CH4 16.04 g/mol; CI2 70.90 g/mol; HCI 36.46 g/mol. a) Write a balanced equation for this irreversible reaction. b) If 33.62 g of methane and 151.76 g of chlorine were available, how many mole or each reagent is this equivalent to? c) Which reagent is the limiting reagent? Explain. d) Calculate the theoretical yield of chloroform in grams. e) If 65.0 g of chloroform actually forms, what is the percentage yield of this reaction? (Pis no more than a page long answer)

Explanation / Answer

a) A balanced chemical equation for formation of CHCl3 from CH4 and Cl2 is

   CH4 + 3Cl2    ------------> CHCl3 + 3HCl

Methane    Chlorine gas           Chloroform

Reaction stoichiometry says,

1 mole of CH4 3 moles of Cl2 1 mole of CHCl3 3 moles of HCl

With given molar masses we can write,

16.04 g mole of CH4 3 x 70.90 of Cl2 119.4 g of CHCl3 3x36.46 g of HCl

i.e. 16.04 g mole of CH4 212.70 g of Cl2 119.4 g of CHCl3 109.38 g of HCl

b)

Actually used masses of reactant CH4 and Cl2 are ,

Mass of CH4 = 33.62 g and Mass of Cl2 = 151.76 g

Number of moles of CH4 = Given mass of CH4 /Molar mass of CH4 = 33.62/16.04 = 2.096

Number of moles of Cl2 = 151.76/70.90 = 2.140

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c) From stoichiometry,

If        1 mole of CH4 3 moles of Cl2.

Then, 2.096 moles of CH4 say ‘A’ moles of Cl2

A = 2.096 x 3 = 6.288 moles of Cl2.

But actually only 2.140 moles of Cl2 used hence Cl2 is a limiting reagent.

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d) As Cl2 is a limiting reagent theoretical yield needed to be calculated with respect to Cl2 amount actually used.

Mass of Cl2 used = 151.76 g

From stoichiometry,

If         212.70 g of Cl2 119.4 g of CHCl3

Then, 151.76 g of Cl2 sat ‘MT‘ g of CHCl3

MT x 212.70 = 151.76 x 119.4

MT = 151.76 x 119.4 / 212.70

MT = 85.2 g

Theoretical yield of CHCl3 is 85.2 g.

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e) Actual yield of CHCl3 = 65.0 g

Then,

% Yield of CHCl3 = {(Actual yield of CHCl3)/(Theoretical yield of CHCl3)} x 100

% Yield of CHCl3 = (65.0/85.2) x 100

% Yield of CHCl3 = 76.3 %

Yield of Chloroform is 76.3%.

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