Question

which represents a balanced chemical equation? *
\\( \mathrm{co_2 + h_2o \
ightarrow ch_4 + o_2} \\)
\\( \mathrm{ch_4 + o_2 \
ightarrow co_2 + h_2o} \\)
\\( \mathrm{ch_4 + 2o_2 \
ightarrow co_2 + 2h_2o} \\)
\\( \mathrm{co_2 + 2h_2o \
ightarrow 2ch_4 + 2o_2} \\)

Explanation:

Step1: Recall Balancing Rules

A balanced chemical equation has equal number of each atom on both sides. We check C, H, O atoms for each option.

Step2: Analyze Option 1 ($\ce{CO_{2} + H_{2}O -> CH_{4} + O_{2}}$)

• C: Left=1, Right=1 (ok)
• H: Left=2, Right=4 (not ok)
• O: Left=3, Right=2 (not ok) → Unbalanced.

Step3: Analyze Option 2 ($\ce{CH_{4} + O_{2} -> CO_{2} + H_{2}O}$)

• C: Left=1, Right=1 (ok)
• H: Left=4, Right=2 (not ok)
• O: Left=2, Right=3 (not ok) → Unbalanced.

Step4: Analyze Option 3 ($\ce{CH_{4} + 2O_{2} -> CO_{2} + 2H_{2}O}$)

• C: Left=1, Right=1 (ok)
• H: Left=4, Right=2×2=4 (ok)
• O: Left=2×2=4, Right=2 + 2×1=4 (ok) → Balanced.

Step5: Analyze Option 4 ($\ce{CO_{2} + 2H_{2}O -> 2CH_{4} + 2O_{2}}$)

• C: Left=1, Right=2 (not ok)
• H: Left=4, Right=8 (not ok)
• O: Left=4, Right=4 (ok) → Unbalanced.

Answer:

$\ce{CH_{4} + 2O_{2} -> CO_{2} + 2H_{2}O}$ (the third option: $\boldsymbol{\ce{CH_{4} + 2O_{2} -> CO_{2} + 2H_{2}O}}$)