What is the product of the unbalanced combustion reaction below? C4H10(g) + O2(g) →

Chemistry

1 answer:

Afina-wow [57]2 years ago

5 0

Answer:

Option C . CO2(g) + H2O(g)

Explanation:

When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

Thus, the product of the unbalanced combustion reaction is:

CO2(g) + H2O(g)

Thus, we can balance the equation as follow:

C2H4(g) + O2(g) —› CO2(g) + H2O(g)

There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:

C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)

There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:

C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)

Thus, the equation is balanced.

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2 years ago

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Answer:

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Explanation:

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Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

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