Question

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. (Use the lowest possible whole number coefficients. Include states-of-matter under SATP conditions in your answer.) C₃H₈(g) + O₂(g) $\longrightarrow$ CO₂(g) + H₂O(l)

Answer 1

Answer:

Balanced reaction: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Reduced: O₂

Oxidized : C₃H₈

Explanation:

For the given reaction:

C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l)

In the reactants, there are 3 C, and in the products only one, so we multiply CO₂ by 3:

C₃H₈(g) + O₂(g) → 3CO₂(g) + H₂O(l)

In the reactants, there are 8 H, and in the products, there are 2 H, so we multiply H₂O by 4:

C₃H₈(g) + O₂(g) → 3CO₂(g) + 4H₂O(l)

In the reactants are 2 O, and in the products 10 O, so we multiply O₂ by 5:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The reaction is balanced.

Let's identify the oxidation number (Nox) of the elements in each compound.

C₃H₈:

H has fix Nox equal to +1, and the molecule is neutral, so, calling x the Nox of C:

3x + 8 = 0

3x = -8

x = -8/3

O₂:

Because is a pure compound, the Nox of O is 0.

CO₂:

The Nox of O is fix equal to -2, and the molecule is neutral, so, calling x the Nox of C:

2x -4 = 0

2x = 4

x = +2

H₂O:

The Nox of H and O are fixed, respectively, +1 and -2.

So, the carbon in C₃H₈ is oxidized because its Nox is increasing, and oxygen in O₂ is reduced because its Nox is decreasing.