59 gram of an amide obtained from a carboxylic acid (RCOOH) upon heating with alkali liberated 17g of `NH_(3)` The acid is :

Options

A. Formic acid
B. Acetic acid
C. Propionic acid
D. Benzoic acid

Explanation

To solve the problem, we need to identify the carboxylic acid (RCOOH) from which the amide (RCONH2) is derived. Given that 59 grams of the amide liberates 17 grams of ammonia (NH3) upon heating with alkali, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Molar Mass of Ammonia:**
The molar mass of ammonia (NH3) is calculated as follows:
- Nitrogen (N) = 14 g/mol
- Hydrogen (H) = 1 g/mol × 3 = 3 g/mol
- Total = 14 + 3 = 17 g/mol

2. **Determine the Moles of Ammonia Produced:**
Since 17 grams of ammonia corresponds to 1 mole (as the molar mass of NH3 is 17 g/mol), we conclude that 1 mole of the amide (RCONH2) will produce 1 mole of ammonia.

3. **Calculate the Molar Mass of the Amide:**
We know that the mass of the amide is 59 grams. Therefore, the molar mass of the amide (RCONH2) is 59 g/mol.

4. **Set Up the Equation for the Molar Mass of the Amide:**
The molar mass of the amide can be expressed as:
\[ \text{Molar mass of amide} = R + 12 + 16 + 2 = R + 30 \]
where R is the molecular weight of the hydrocarbon part of the amide.

5. **Solve for R:**
We know that the molar mass of the amide is 59 g/mol, so we can set up the equation:
\[ R + 30 = 59 \]
Solving for R gives:
\[ R = 59 - 30 = 29 \]

6. **Identify the Hydrocarbon Group (R):**
The molecular weight of R is 29 g/mol. The possible structure for R that fits this weight is:
- R = CH2CH3 (ethyl group), which has a molar mass of 29 g/mol.

7. **Determine the Corresponding Carboxylic Acid:**
The carboxylic acid corresponding to the amide RCONH2 (where R = CH2CH3) is:
\[ \text{Carboxylic acid} = RCOOH = CH3CH2COOH \]
This is propanoic acid (C2H5COOH).

### Conclusion:
The carboxylic acid from which the amide is derived is **propanoic acid (C2H5COOH)**.