The primary alcohol 1 - Propanol can be oxidized with Aqueous nitric acid `HNO_(3)` are required for complete oxidation of 180 gm of 1-Propanol ?
`CH_(3)CH_(2)OH+HNO_(3) rarr CH_(3)CH_(2)COOH+NO+H_(2)O`

Explanation

To find out how many moles of aqueous nitric acid (HNO3) are required for the complete oxidation of 180 g of 1-Propanol (C3H8O), we will follow these steps:

### Step 1: Determine the Molar Mass of 1-Propanol
1-Propanol has the molecular formula C3H8O. We can calculate its molar mass as follows:
- Carbon (C): 12.01 g/mol × 3 = 36.03 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol

Adding these together:
\[ \text{Molar mass of 1-Propanol} = 36.03 + 8.064 + 16.00 = 60.094 \text{ g/mol} \approx 60 \text{ g/mol} \]

### Step 2: Calculate the Number of Moles of 1-Propanol
Using the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]
Substituting the values:
\[ \text{Number of moles of 1-Propanol} = \frac{180 \text{ g}}{60 \text{ g/mol}} = 3 \text{ moles} \]

### Step 3: Write the Balanced Chemical Equation
The unbalanced reaction is given as:
\[ \text{C3H8O} + \text{HNO3} \rightarrow \text{C3H6O2} + \text{NO} + \text{H2O} \]
To balance it, we need to consider the oxidation states:
- The oxidation state of carbon in 1-Propanol (C3H8O) changes from -1 to +3 in propanoic acid (C3H6O2).
- The oxidation state of nitrogen in HNO3 changes from +5 to +2.

Balancing the changes in oxidation states, we find:
\[ 3 \text{C3H8O} + 4 \text{HNO3} \rightarrow 3 \text{C3H6O2} + 4 \text{NO} + 5 \text{H2O} \]

### Step 4: Determine the Moles of HNO3 Required
From the balanced equation, we see that:
- 3 moles of 1-Propanol require 4 moles of HNO3 for complete oxidation.

Thus, for 3 moles of 1-Propanol, we need:
\[ \text{Moles of HNO3 required} = 4 \text{ moles} \]

### Final Answer
The number of moles of HNO3 required for the complete oxidation of 180 g of 1-Propanol is **4 moles**.

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