acetic acid (hc₂h₃o₂) is the primary component of vinegar. it is produced along with water (h₂o) when methanol (ch₃oh) reacts with carbon dioxide (co₂) and hydrogen gas (h₂). which chemical equation accurately represents this reaction? ch₃oh (l) + co₂ (g) + h₂ (g) → hc₂h₃o₂ (l) + h₂o (l) ch₃oh (l) + 3co₂ (g) + 5h₂ (g) → 3hc₂h₃o₂ (l) + h₂o (l) 2ch₃oh (l) + 2co₂ (g) + h₂ (g) → 2hc₂h₃o₂ (l) + h₂o (l) 2ch₃oh (l) + co₂ (g) + 2h₂ (g) → 2hc₂h₃o₂ (l) + 2h₂o (l)

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Accepted Answer

# Explanation:
## Step1: Count atoms on both sides
For carbon ($C$):
- In $2CH_{3}OH$, there are $2\times1 = 2$ $C$ atoms.
- In $CO_{2}$, there is $1$ $C$ atom. So total $C$ atoms on reactant side: $2 + 1=3$.
- In $2HC_{2}H_{3}O_{2}$, there are $2\times2 = 4$ $C$ atoms. This is incorrect.
For hydrogen ($H$):
- In $2CH_{3}OH$, there are $2\times4 = 8$ $H$ atoms.
- In $2H_{2}$, there are $2\times2 = 4$ $H$ atoms. So total $H$ atoms on reactant side: $8+4 = 12$.
- In $2HC_{2}H_{3}O_{2}$, there are $2\times4 = 8$ $H$ atoms. In $2H_{2}O$, there are $2\times2 = 4$ $H$ atoms. Total $H$ atoms on product side: $8 + 4=12$.
For oxygen ($O$):
- In $2CH_{3}OH$, there are $2\times1 = 2$ $O$ atoms.
- In $CO_{2}$, there are $2$ $O$ atoms. So total $O$ atoms on reactant side: $2+2 = 4$.
- In $2HC_{2}H_{3}O_{2}$, there are $2\times2 = 4$ $O$ atoms. In $2H_{2}O$, there are $2\times1 = 2$ $O$ atoms. Total $O$ atoms on product side: $4+2 = 6$. This is incorrect.

## Step2: Check another option
For $2CH_{3}OH(l)+2CO_{2}(g)+H_{2}(g)\to2HC_{2}H_{3}O_{2}(l)+H_{2}O(l)$
- **Carbon ($C$)**:
  - Reactant side: In $2CH_{3}OH$, $2\times1 = 2$ $C$ atoms. In $2CO_{2}$, $2\times1 = 2$ $C$ atoms. Total $C$ atoms: $2 + 2=4$.
  - Product side: In $2HC_{2}H_{3}O_{2}$, $2\times2 = 4$ $C$ atoms.
- **Hydrogen ($H$)**:
  - Reactant side: In $2CH_{3}OH$, $2\times4 = 8$ $H$ atoms. In $H_{2}$, $2$ $H$ atoms. Total $H$ atoms: $8+2 = 10$.
  - Product side: In $2HC_{2}H_{3}O_{2}$, $2\times4 = 8$ $H$ atoms. In $H_{2}O$, $2$ $H$ atoms. Total $H$ atoms: $8 + 2=10$.
- **Oxygen ($O$)**:
  - Reactant side: In $2CH_{3}OH$, $2\times1 = 2$ $O$ atoms. In $2CO_{2}$, $2\times2 = 4$ $O$ atoms. Total $O$ atoms: $2+4 = 6$.
  - Product side: In $2HC_{2}H_{3}O_{2}$, $2\times2 = 4$ $O$ atoms. In $H_{2}O$, $1$ $O$ atom. Total $O$ atoms: $4 + 1=5$. This is incorrect.

## Step3: Check $CH_{3}OH(l)+3CO_{2}(g)+5H_{2}(g)\to3HC_{2}H_{3}O_{2}(l)+H_{2}O(l)$
- **Carbon ($C$)**:
  - Reactant side: In $CH_{3}OH$, $1$ $C$ atom. In $3CO_{2}$, $3\times1 = 3$ $C$ atoms. Total $C$ atoms: $1+3 = 4$.
  - Product side: In $3HC_{2}H_{3}O_{2}$, $3\times2 = 6$ $C$ atoms. This is incorrect.

## Step4: Check $CH_{3}OH(l)+CO_{2}(g)+H_{2}(g)\to HC_{2}H_{3}O_{2}(l)+H_{2}O(l)$
- **Carbon ($C$)**:
  - Reactant side: In $CH_{3}OH$, $1$ $C$ atom. In $CO_{2}$, $1$ $C$ atom. Total $C$ atoms: $1 + 1=2$.
  - Product side: In $HC_{2}H_{3}O_{2}$, $2$ $C$ atoms.
- **Hydrogen ($H$)**:
  - Reactant side: In $CH_{3}OH$, $4$ $H$ atoms. In $H_{2}$, $2$ $H$ atoms. Total $H$ atoms: $4+2 = 6$.
  - Product side: In $HC_{2}H_{3}O_{2}$, $4$ $H$ atoms. In $H_{2}O$, $2$ $H$ atoms. Total $H$ atoms: $4 + 2=6$.
- **Oxygen ($O$)**:
  - Reactant side: In $CH_{3}OH$, $1$ $O$ atom. In $CO_{2}$, $2$ $O$ atoms. Total $O$ atoms: $1+2 = 3$.
  - Product side: In $HC_{2}H_{3}O_{2}$, $2$ $O$ atoms. In $H_{2}O$, $1$ $O$ atom. Total $O$ atoms: $2+1 = 3$.

# Answer:
$CH_{3}OH(l)+CO_{2}(g)+H_{2}(g)\to HC_{2}H_{3}O_{2}(l)+H_{2}O(l)$

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Explanation:

Step1: Count atoms on both sides

Step2: Check another option

Step3: Check \(CH_{3}OH(l)+3CO_{2}(g)+5H_{2}(g)\to3HC_{2}H_{3}O_{2}(l)+H_{2}O(l)\)

Step4: Check \(CH_{3}OH(l)+CO_{2}(g)+H_{2}(g)\to HC_{2}H_{3}O_{2}(l)+H_{2}O(l)\)

Answer: