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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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Answer:

V = 364500 L, 476.748 yard³

Explanation:

Given that,

The dimensions of a room are 10 meters wide by 15 meters long and 8.0 ft high.

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The volume of the room is :

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V = 10×15×2.43

V = 364.5 m³

As 1 m³ = 1000 L

364.5 m³ = 364500 L

Also, 1 m³ = 1.30795 yard³

364.5 m³ = 476.748 yard³

Hence, this is the required solution.

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3 years ago
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3 years ago
How do you work out question 1a?
Sliva [168]

Answer:

-125 kJ

Explanation:

You calculate the energy required to break all the bonds in the reactants. Then you subtract the energy to break all the bonds in the products.

                     H₂C=CH₂   +    H₂ ⟶    H₃C-CH₃

Bonds:       4C-H + 1C=C     1H-H     6C-H + 1C-C

D/kJ·mol⁻¹:  413       612        436       413      347

The formula relating ΔHrxn and bond dissociation energies (D) is

ΔHrxn = Σ(Dreactants) – Σ(Dproducts)

(Note: This is an exception to the rule. All other thermochemical reactions are “products – reactants”. With bond energies, it’s “reactants – products”. The reason comes from the way we define bond energies.)

<em>For the reactant</em>s:

Σ(Dreactants) = 4 × 413 + 1 × 612 + 1 × 436 = 2700 kJ

<em>For the products:</em>

Σ(Dproducts) = 6 × 413 + 1 × 347 = 2825 kJ

<em>For the system</em> :

ΔHrxn = 2700 - 2825 = -125 kJ

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