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Liono4ka [1.6K]
2 years ago
5

How many moles of oxygen are consumed in the complete combustion of 1.60 moles of benzene, c6h6?

Chemistry
1 answer:
My name is Ann [436]2 years ago
3 0

In the complete combustion of 1.60 moles of benzene, C6H6, 12 moles of oxygen, O2, is consumed.

Combustion is defined as the process of burning something. In chemistry, combustion refers to the chemical process between a fuel and an oxidant, usually oxygen to produce heat and light in the form of flame.

In a complete combustion, oxygen is sufficient to react with any hydrocarbons to produce carbon dioxide and water.

Balancing the combustion reaction of benzene, we have:

2C6H6 + 15 O2 = 12CO2 + 6H2O

Based on the balanced combustion reaction above, 2 moles of benzene requires 15 moles of oxygen to have a complete combustion.

If we have 1.60 moles C6H6,

moles O2 = mole ratio x mole of benzene

moles O2 = (15 moles O2/2 moles C6H6) x 1.60 moles C6H6

moles O2 = 12

To learn more about combustion: brainly.com/question/9913173

#SPJ4

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d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

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From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = \frac{ 815.74}{44.1 }

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of  1,006.29 grams of oxygen =\frac{ 1,006.29}{32 }

Mole of  1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

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v =  3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane (C_3H_8) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

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