1. **Stating the problem:** We need to find the product of the reaction between propanoic acid and methanol in the presence of sulfuric acid (H2SO4). 2. **Reaction type:** This is an esterification reaction, where a carboxylic acid reacts with an alcohol in the presence of an acid catalyst to form an ester and water. 3. **General formula:** The reaction can be represented as: $$\text{R-COOH} + \text{R'-OH} \xrightarrow{H_2SO_4} \text{R-COOR'} + H_2O$$ where R-COOH is the carboxylic acid and R'-OH is the alcohol. 4. **Applying to our reactants:** Propanoic acid is $\mathrm{CH_3CH_2COOH}$ and methanol is $\mathrm{CH_3OH}$. 5. **Product formation:** The ester formed is methyl propanoate: $$\mathrm{CH_3CH_2COOH} + \mathrm{CH_3OH} \xrightarrow{H_2SO_4} \mathrm{CH_3CH_2COOCH_3} + H_2O$$ 6. **Explanation:** The hydroxyl group ($-OH$) from the acid and a hydrogen ($H$) from the alcohol combine to form water, and the remaining parts join to form the ester bond. **Final answer:** The product is methyl propanoate ($\mathrm{CH_3CH_2COOCH_3}$) and water.