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2 years ago

The by-product of the chlorination of an alkane is

Chemistry

1 answer:

Elanso [62]2 years ago

7 0

The by-product of the chlorination of an alkane is <u>HCl</u>

Explanation:

Chlorination is the process of adding chlorine to drinking water to disinfect it and kill germs. Different processes can be used to achieve safe levels of chlorine in drinking water.
Chlorination of alkane gives a mixture of different products.
When consider mechanism of alkanes chlorination, free radicals are formed during the reaction to keep the continuous reaction.
Different alkyl chloride compounds, extended carbon chains compounds and HCl are formed as products in product mixture.
Chlorination byproducts, their toxicodynamics and removal from drinking water.
Halogenated trihalomethanes (THMs) and haloacetic acids (HAAs) are two major classes of disinfection byproducts (DBPs) commonly found in waters disinfected with chlorine
Chlorine is available as compressed elemental gas, sodium hypochlorite solution (NaOCl) or solid calcium hypochlorite (Ca(OCl)2

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An unknown liquid is composed of 34.31% c, 5.28% h, and 60.41% i. The molecular weight is 210.06 amu. What is the molecular form

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In an unknown liquid, the percentage composition with respect to carbon, hydrogen and iodine is 34.31%, 5.28% and 60.41% respectively.

Let the mass of liquid be 100 g thus, mass of carbon, hydrogen and oxygen will be 34.31 g, 5.28 g and 60.41 g respectively.

To calculate molecular formula of compound, convert mass into number of moles as follows:

$n=\frac{m}{M}$

Molar mass of carbon, hydrogen and iodine is 12 g/mol, 1 g/mol and 126.90 g/mol.

Taking the ratio:

$C:H:I=n_{C}:n_{H}:n_{I}$

Putting the values,

$C:H:I=\frac{34.31 g}{12 g/mol}:\frac{5.28 g}{1 g/mol}:\frac{60.41 g}{126.90 g/mol}=6:11:1$

Thus, molecular formula of compound will be $C_{6}H_{11}I$ .

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The by-product of the chlorination of an alkane is​

The by-product of the chlorination of an alkane is