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3 years ago

How many moles of O2 are required to generate 18 moles of H2O in the given reaction? 2C8H18 + 25O2 16CO2 + 18H2O

Chemistry

2 answers:

faltersainse [42]3 years ago

8 0

Answer: 25 moles of oxygen gas will be required for the production of given amount of water.

Explanation:

This is a reaction of combustion of hydrocarbon. Combustion reaction states that when a hydrocarbon reacts with excess of oxygen it results in the formation of carbon dioxide and water.

For the reaction of combustion of hydrocarbon, the equation follows:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

By Stoichiometry of the reaction:

18 moles of water is produced by 25 moles of oxygen gas.

So, 18 moles of water will be produced from = $\frac{25}{18}\times 18=25mol$ of oxygen gas.

Hence, 25 moles of oxygen gas will be required for the production of given amount of water.

Veronika [31]3 years ago

7 0

There's a slight error in your equation. I think you were trying to present it like this:

2C8H18 + 25O2 -> 16CO2 + 18H2O

Mole Ratio
O2 : H20
25 : 18
? moles : 18 moles
(18/18)×25 : 18 moles

25 moles : 18 moles

Final answer would be 25 moles of O2. :)

If you have any doubts that you want to clarify with me, please ask me! :)
I will do my utmost best to help you.

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3 years ago

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The total pressure of the mixture in the tank of volume 6.25 litres at 51°C is 1291.85 kPa.

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Pressure(P₁)=125 kPa

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Pressure(P₂)= 125 kPa

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Then, for the mixture,

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Then, By Combined gas laws,

$\frac{P_{1} V_{1} }{T_{1} } +\frac{P_{2} V_{2} }{T_{2} } =\frac{PV}{T}$

or, $\frac{15.1*125}{298} +\frac{44.3*125}{298} =\frac{P*6.25}{324}$

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