The reaction of methane (CH_4) with Cl_2 forms chloroform (CHCl_3), and HC1. Alt

Question

The reaction of methane (CH_4) with Cl_2 forms chloroform (CHCl_3), and HC1. Although CHCl_3 is a general anesthetic, it is no longer used for this purpose since it is also carcinogenic. The molar masses for all substances are given under the balanced equation. CH_4(g) + 3 Cl_2(g) rightarrow CHCl_3(l) + 3HCl(g) 16.04 g/mol 70.90 g/mol 119.4 g/mol 36.46 g/mol a. What is the theoretical yield of CHCl_3 in grams from 7.40 g of CH_4? b. What is the percent yield if 27.5 g of CHCl_3, are actually formed in this reaction?

Explanation / Answer

a)

molar mass of CH4 = 16.04 g/mol

number of mole of CH4 = given mass/molar mass

number of mole of CH4 = 7.4/16.04

number of mole of CH4 = 0.46 mole

balanced equation is

CH4 + 3Cl2 --> CHCl3 + 3HCl

according to given chemical equation

number of CH4 used =number of mole of CHCl3 formed

mole of CHCl3 formed = 0.46 mole

mass of CHCl3 formed = mole*molar mass

mass of CHCl3 formed =(0.46*119.4)

mass of CHCl3 formed = 54.9 g

b)

But experimental yield of CHCl3 given is 27.5 g

therotical yield of CHCl3 = 54.9 g

% yield = (experimental yield/ therotical yield)*100

% yield of CHCl3 = ( 27.5/54.9)*100

% yield of CHCl3 = 50.1 %

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