Final answer:
When methanol (CH₃OH) is treated with potassium hydroxide (KOH), it forms potassium methoxide (CH₃OK) and water (H₂O).
Step-by-step explanation:
The correct answer is:
A) CH₃OH+KOH→CH₃OK+H₂O
When methanol (CH₃OH) is treated with potassium hydroxide (KOH), it forms potassium methoxide (CH₃OK) and water (H₂O). This reaction is an example of a neutralization reaction between an acid (methanol) and a base (potassium hydroxide).
Overall, the reaction can be written as:
CH₃OH + KOH → CH₃OK + H₂O