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2 years ago

Write and balance the equation for the combustion of the fatty acid lauric acid, (C12H24O2).

Chemistry

1 answer:

Harlamova29_29 [7]2 years ago

4 0

C12H24O2 +17 O2-------->12CO2 + 12H2O

<h3>Combustion:-</h3>

combustion is a chemical reaction that often involves the presence of oxygen and produces heat and light in the form of flames.

<h3>Lauric acid:-</h3>

Lauric acid has a 12-carbon backbone and is a saturated medium-chain fatty acid. In addition to being a key component of coconut oil and palm kernel oil, lauric acid occurs naturally in a variety of plant and animal fats and oils.

White solid lauric acid has a little bay oil odour to it.

Lauric acid is a cheap, non-toxic, and easy-to-handle substance that is frequently employed in lab studies on melting-point depression. Because lauric acid is a solid at ambient temperature but a liquid at boiling temperatures, it can be used to test different solutes to determine their molecular weights.

To learn more about Fatty acids refer to :-

brainly.com/question/26353151

#SPJ10

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3 years ago

There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

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Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

$CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ$ ...[1]

Step 2 :

$6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ$ ..[2]

Adding 6 × [1] and [2]:

$6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)$

$6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

we get :

$6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?$

$\Delta H'=6\times \Delta H_1+\Delta H_2$

$\Delta H'=6\times 414.0 kJ+132.0kJ$

$\Delta H'=2,626 kJ$

Energy released on formation of 5 moles of acrylic acid = 2,626 kJ

Energy released on formation of 1 mole of acrylic acid:

$\frac{ 2,626 kJ}{5 } = 523.2 kJ$

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