How many grams of caustic potash required to completely neutralise `12.6 g HNO_(3)`?

Options

A. (a)`22.4 KOH `
B. (b)` 1.01 KOH`
C. (c )` 6.02 KOH`
D. (d)`11.2 KOH `

Explanation

To determine how many grams of caustic potash (KOH) are required to completely neutralize 12.6 grams of nitric acid (HNO3), we can follow these steps:

### Step 1: Write the balanced chemical equation for the reaction.
The neutralization reaction between KOH and HNO3 can be represented as:
\[ \text{KOH} + \text{HNO}_3 \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of HNO3.
To find the molar mass of HNO3, we add the atomic masses of its constituent elements:
- Hydrogen (H): 1 g/mol
- Nitrogen (N): 14 g/mol
- Oxygen (O): 16 g/mol (and there are 3 oxygen atoms)

So, the molar mass of HNO3 is:
\[ 1 + 14 + (3 \times 16) = 1 + 14 + 48 = 63 \, \text{g/mol} \]

### Step 3: Calculate the number of moles of HNO3 in 12.6 grams.
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
we can calculate the moles of HNO3:
\[ \text{Number of moles of HNO}_3 = \frac{12.6 \, \text{g}}{63 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 4: Determine the moles of KOH required.
From the balanced equation, we see that the mole ratio of KOH to HNO3 is 1:1. Therefore, the number of moles of KOH required is also 0.2 mol.

### Step 5: Calculate the molar mass of KOH.
Now we calculate the molar mass of KOH:
- Potassium (K): 39 g/mol
- Oxygen (O): 16 g/mol
- Hydrogen (H): 1 g/mol

So, the molar mass of KOH is:
\[ 39 + 16 + 1 = 56 \, \text{g/mol} \]

### Step 6: Calculate the mass of KOH required.
Using the number of moles of KOH and its molar mass, we can find the mass required:
\[ \text{Mass of KOH} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of KOH} = 0.2 \, \text{mol} \times 56 \, \text{g/mol} = 11.2 \, \text{g} \]

### Final Answer:
The mass of caustic potash (KOH) required to completely neutralize 12.6 grams of HNO3 is **11.2 grams**.

---