Calcium hydroxide and ammonium chloride react to give ammonia as per equation:
`Ca(OH)_(2) + 2 NH_(4)Cl to CaCl_(2) + 2 NH_(3) + 2 H_(2) O `
In a reaction, `5.35 `g of ammonium chloride were consumed . Calculate
The volume at STP of ammonia liberated.
`(R.A.M.: H = 1, N = 14 , O = 16, Cl = 35.5, Ca = 40) `

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the molar mass of Ammonium Chloride (NH₄Cl)

The formula for calculating the molar mass is:
\[ \text{Molar Mass} = \text{mass of N} + \text{mass of H} + \text{mass of Cl} \]

Using the given atomic masses:
- Nitrogen (N) = 14 g/mol
- Hydrogen (H) = 1 g/mol (4 H atoms in NH₄Cl)
- Chlorine (Cl) = 35.5 g/mol

Calculating the molar mass:
\[ \text{Molar Mass of NH₄Cl} = 14 + (4 \times 1) + 35.5 = 14 + 4 + 35.5 = 53.5 \text{ g/mol} \]

### Step 2: Determine the number of moles of Ammonium Chloride (NH₄Cl) consumed

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given mass of NH₄Cl = 5.35 g, we calculate:
\[ \text{Number of moles of NH₄Cl} = \frac{5.35 \text{ g}}{53.5 \text{ g/mol}} \approx 0.1 \text{ moles} \]

### Step 3: Use the stoichiometry of the reaction to find moles of Ammonia (NH₃) produced

From the balanced equation:
\[ \text{Ca(OH)}_2 + 2 \text{NH}_4\text{Cl} \rightarrow \text{CaCl}_2 + 2 \text{NH}_3 + 2 \text{H}_2\text{O} \]
We see that 2 moles of NH₄Cl produce 2 moles of NH₃. Thus, 1 mole of NH₄Cl produces 1 mole of NH₃.

Since we have 0.1 moles of NH₄Cl, it will produce:
\[ \text{Moles of NH₃} = 0.1 \text{ moles} \]

### Step 4: Calculate the volume of Ammonia (NH₃) at STP

At Standard Temperature and Pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of 0.1 moles of NH₃ is:
\[ \text{Volume of NH₃} = 0.1 \text{ moles} \times 22.4 \text{ L/mol} = 2.24 \text{ L} \]

### Final Answer:
The volume of ammonia liberated at STP is **2.24 liters**.

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