How many grams of oxygen are needed to complete the combustion reaction of 4.80 moles of pentane (C5H12)?

1228.8 g of O₂ are needed in the reaction

Explanation:

Combustions reactions make CO₂ and H₂O as products. Let's make the reaciton:

C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O

Oxygen is always one of the reactants

You have 4.80 moles of pentane. Let's make the rule of three (ratio is 1:8)

1 moles of pentane needs 8 moles of oxygen to react

Therefore 4.80 moles will react with (4.80 . 8) /1 = 38.4 moles

We convert the moles to mass: 38.4 mol . 32 g / 1mol = 1228.8 g of O₂

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

i would say b

explanation:

but i think i would need more information than that