Question

methane (ch₄) and oxygen (o₂) combine to make carbon dioxide (co₂) and water (h₂o) in the following unbalanced equation.

ch₄ + o₂ → co₂ + h₂o

how many oxygen molecules are necessary in order to complete this reaction?

(1 point)

two

one

four

three

Explanation:

Step1: Balance the carbon atoms

In the unbalanced equation \( \text{CH}_4 + \text{O}_2 ightarrow \text{CO}_2 + \text{H}_2\text{O} \), the number of carbon atoms on the left (in \( \text{CH}_4 \)) is 1, and on the right (in \( \text{CO}_2 \)) is also 1. So carbon is balanced for now.

Step2: Balance the hydrogen atoms

On the left, \( \text{CH}_4 \) has 4 hydrogen atoms. On the right, \( \text{H}_2\text{O} \) has 2 hydrogen atoms per molecule. To balance hydrogen, we need 2 molecules of \( \text{H}_2\text{O} \) (since \( 2 \times 2 = 4 \) hydrogen atoms). Now the equation becomes \( \text{CH}_4 + \text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \).

Step3: Balance the oxygen atoms

On the right, \( \text{CO}_2 \) has 2 oxygen atoms and \( 2\text{H}_2\text{O} \) has \( 2 \times 1 = 2 \) oxygen atoms, so total oxygen atoms on the right are \( 2 + 2 = 4 \). On the left, \( \text{O}_2 \) has 2 oxygen atoms per molecule. To get 4 oxygen atoms, we need \( \frac{4}{2} = 2 \) molecules of \( \text{O}_2 \)? Wait, no, wait. Wait, let's re - check. Wait, the balanced equation for the combustion of methane is \( \text{CH}_4 + 2\text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)? No, wait, no. Wait, let's do it properly.

Wait, \( \text{CH}_4 \) (1 C, 4 H), \( \text{O}_2 \) (O), \( \text{CO}_2 \) (1 C, 2 O), \( \text{H}_2\text{O} \) (2 H, 1 O).

Balance C: already 1 on each side.

Balance H: 4 H on left, so we need 2 \( \text{H}_2\text{O} \) (since 2*2 = 4 H). Now the right side has \( \text{CO}_2 + 2\text{H}_2\text{O} \), which has 2 O (from \( \text{CO}_2 \)) + 2 O (from 2 \( \text{H}_2\text{O} \)) = 4 O.

On the left, \( \text{O}_2 \) has 2 O per molecule. So to get 4 O, we need 2 \( \text{O}_2 \) molecules? Wait, no, wait, the correct balanced equation is \( \text{CH}_4 + 2\text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)? Wait, no, that gives 4 O on left (2*2) and 4 O on right (2 + 2). But wait, the question is about the number of \( \text{O}_2 \) molecules. Wait, maybe I made a mistake. Wait, let's re - balance:

1. Start with \( \text{CH}_4 + \text{O}_2 ightarrow \text{CO}_2 + \text{H}_2\text{O} \)

2. Balance C: 1 on left, 1 on right (good).

3. Balance H: 4 on left (\( \text{CH}_4 \)), so we need 2 \( \text{H}_2\text{O} \) (since 2*2 = 4 H). Now equation: \( \text{CH}_4 + \text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)

4. Balance O: On right, \( \text{CO}_2 \) has 2 O, \( 2\text{H}_2\text{O} \) has 2 O (2*1), total 4 O. On left, \( \text{O}_2 \) has 2 O per molecule. So number of \( \text{O}_2 \) molecules needed: \( \frac{4}{2}=2 \)? Wait, but that's not right. Wait, no, the correct balanced equation for methane combustion is \( \text{CH}_4 + 2\text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)? Wait, no, 2 \( \text{O}_2 \) is 4 O atoms, and on the right, \( \text{CO}_2 \) (2 O) + 2 \( \text{H}_2\text{O} \) (2 O) = 4 O. So yes, 2 \( \text{O}_2 \) molecules? But wait, the options have "two", "one", "four", "three". Wait, maybe I messed up. Wait, no, let's check again.

Wait, \( \text{CH}_4 + \text{O}_2 ightarrow \text{CO}_2 + \text{H}_2\text{O} \)

After balancing H: \( \text{CH}_4 + \text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)

Now O on right: 2 (from \( \text{CO}_2 \)) + 2 (from 2 \( \text{H}_2\text{O} \)) = 4. So O on left: \( \text{O}_2 \) has 2 per molecule. So number of \( \text{O}_2 \) molecules: 4/2 = 2? But wait, the correct balanced equation is \( \text{CH}_4 + 2\text{O}_2 ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)? Wait, no, 2 \( \text{O}_2 \) is 4 O, and right side is 2 + 2 = 4 O. So the number of \( \text{O}_2 \) molecules needed is 2? But wait, the options include "two" as an option. Wait, but maybe I made a mistake. Wait, let's count the O atoms again.

Wait, in the balanced equation, for 1 \( \text{CH}_4 \), we need 2 \( \text{O}_2 \) to produce 1 \( \text{CO}_2 \) and 2 \( \text{H}_2\text{O} \). So the number of \( \text{O}_2 \) molecules is 2? But wait, the options: two, one, four, three. Wait, maybe the question is about the coefficient of \( \text{O}_2 \) in the balanced equation. Let's re - do the balancing:

- Carbon: 1 on left (CH₄), 1 on right (CO₂) → balanced.

- Hydrogen: 4 on left (CH₄), so we need 2 H₂O (2×2 = 4 H) → now equation: CH₄ + O₂ → CO₂ + 2H₂O.

- Oxygen: On the right, CO₂ has 2 O, 2H₂O has 2 O (2×1) → total 4 O. On the left, O₂ has 2 O per molecule. So to get 4 O, we need 2 O₂ molecules (since 2×2 = 4 O). So the coefficient of O₂ is 2, meaning 2 oxygen molecules are needed.

Answer:

two (the option corresponding to two oxygen molecules)