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NikAS [45]
2 years ago
10

if 4.50 mol of ethane, c_(2)h_(6),undergo combustion according to the equation below , how many moles of oxygen are required? 2C

_(2)H_(6)+7O_(2 )->4CO_(2)+6H_(2)O
Chemistry
1 answer:
alexira [117]2 years ago
4 0

15.75 moles of oxygen is required for 4.5 mole of ethane.

<h3>What is a Combustion Reaction ?</h3>

The reaction in which a fuel  gets oxidized by an oxidizing agent is called a Combustion Reaction , releasing a large amount of energy is called Combustion Reaction.

It is given in the question that

4.50 mole of ethane undergoes Combustion

2C₂H₆   +  7O₂   ->  4CO₂  +6H₂O

Mole fraction of ethane to oxygen is 2:7

Moles of ethane = 4.5

4.5 /x = 2/7

4.5 * 7 / 2 = x

x = 15.75 moles.

Therefore 15.75 moles of oxygen is required for 4.5 mole of ethane.

To know more about Combustion Reaction

brainly.com/question/12172040

#SPJ1

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To balance a chemical equation, both sides should have equal mass, or in other words both sides should have same number of atoms as to follow the conservation of mass rule.

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RHS:

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Also since P₄O₆ has the most number of atoms we will make the LHS equalize to P₄O₆.

Difference between Phosphorous atoms in LHS to RHS = 3

Since phosphorous is a monatomic we need 3 phosphorous atoms extra

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But Oxygen is diatomic, so we need 4/2 = 2 Oxygen molecules

Now lets see if it is balanced

P + 3P + O₂ + 2O₂ -------> P₄O₆

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LHS:

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RHS:

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LHS = RHS

Therefore the balanced equation is 4P + 3O₂ =  P₄O₆

Happy to help :)

If you need more explanation or help in any other question, feel free to ask

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