Question

ch₄ +? o₂ → co₂ + 2 h₂o
what coefficient is needed in front of the o₂?
2 3 4 5

Explanation:

Step1: Balance Carbon Atoms

In the reaction \( \text{CH}_4 +?\text{O}_2
ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \), 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.

Step2: Balance Hydrogen Atoms

The number of hydrogen atoms in \( \text{CH}_4 \) is 4. On the right, in \( 2\text{H}_2\text{O} \), the number of hydrogen atoms is \( 2\times2 = 4 \). So hydrogen is balanced.

Step3: Balance Oxygen Atoms

On the right, the number of oxygen atoms: in \( \text{CO}_2 \) there are 2, and in \( 2\text{H}_2\text{O} \) there are \( 2\times1 = 2 \). So total oxygen on the right is \( 2 + 2 = 4 \). On the left, oxygen is in \( \text{O}_2 \). Let the coefficient of \( \text{O}_2 \) be \( x \). Then the number of oxygen atoms from \( \text{O}_2 \) is \( 2x \). We set \( 2x = 4 \), so \( x = 2 \). Wait, no, wait. Wait, \( \text{CO}_2 \) has 2 O, \( 2\text{H}_2\text{O} \) has 2 O, total O on right is 4? Wait, no: \( \text{CO}_2 \) is 1 molecule with 2 O, \( 2\text{H}_2\text{O} \) is 2 molecules with 1 O each, so total O on right is \( 2 + 2\times1 = 4 \)? Wait, no, \( 2\text{H}_2\text{O} \) has \( 2\times1 = 2 \) O? Wait, no, \( \text{H}_2\text{O} \) has 1 O per molecule, so 2 molecules have 2 O. \( \text{CO}_2 \) has 2 O. So total O on right is \( 2 + 2 = 4 \). On the left, \( \text{O}_2 \) has 2 O per molecule. So to get 4 O, we need \( 4\div2 = 2 \)? Wait, no, wait, let's re - check. Wait, the reaction is \( \text{CH}_4 + \text{O}_2
ightarrow \text{CO}_2 + \text{H}_2\text{O} \) (before balancing). Wait, the given reaction has \( 2\text{H}_2\text{O} \). So let's do it again.

Wait, correct balancing:

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

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

Balance H: 4 on left (\( \text{CH}_4 \)), so on right, \( \text{H}_2\text{O} \) must have 4 H, so \( 2\text{H}_2\text{O} \) (since \( 2\times2 = 4 \) H). Now the equation is \( \text{CH}_4 + \text{O}_2
ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)

Now balance O: on right, \( \text{CO}_2 \) has 2 O, \( 2\text{H}_2\text{O} \) has 2 O, total O is \( 2 + 2 = 4 \). On left, \( \text{O}_2 \) has 2 O per molecule. So number of \( \text{O}_2 \) molecules needed is \( 4\div2 = 2 \). Wait, but that gives \( \text{CH}_4 + 2\text{O}_2
ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \). Let's check O: left has \( 2\times2 = 4 \) O, right has \( 2 + 2 = 4 \) O. H: left \( \text{CH}_4 \) has 4 H, right \( 2\text{H}_2\text{O} \) has 4 H. C: left 1, right 1. So the coefficient of \( \text{O}_2 \) is 2. Wait, but the options are 2,3,4,5. Wait, maybe I made a mistake. Wait, no, let's check again. Wait, \( \text{CH}_4 + 2\text{O}_2
ightarrow \text{CO}_2 + 2\text{H}_2\text{O} \). Yes, that's the balanced equation. So the coefficient of \( \text{O}_2 \) is 2.

Answer:

2