What weight of slaked lime will be required to decompose completely 4 grams ammonium chloride

Options

A. 2.77 g
B. 3.5 g
C. 5.5 g
D. 5.44 g

Answer

To solve the problem of how much slaked lime (calcium hydroxide) is needed to completely decompose 4 grams of ammonium chloride, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between ammonium chloride (NH4Cl) and slaked lime (calcium hydroxide, Ca(OH)2) can be represented as follows:

\[ 2 \, \text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \rightarrow \text{CaCl}_2 + 2 \, \text{NH}_3 + 2 \, \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of ammonium chloride (NH4Cl)
- Nitrogen (N): 14 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol
- Chlorine (Cl): 35.5 g/mol

Total molar mass of NH4Cl:
\[ 14 + 4 + 35.5 = 53.5 \, \text{g/mol} \]

### Step 3: Calculate the molar mass of calcium hydroxide (Ca(OH)2)
- Calcium (Ca): 40 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol

Total molar mass of Ca(OH)2:
\[ 40 + 32 + 2 = 74 \, \text{g/mol} \]

### Step 4: Determine the number of moles of ammonium chloride in 4 grams
Using the molar mass of NH4Cl:
\[ \text{Moles of NH}_4\text{Cl} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{53.5 \, \text{g/mol}} \approx 0.0747 \, \text{mol} \]

### Step 5: Use the stoichiometry of the reaction
From the balanced equation, 2 moles of NH4Cl react with 1 mole of Ca(OH)2. Therefore, the moles of Ca(OH)2 required can be calculated as follows:
\[ \text{Moles of Ca(OH)}_2 = \frac{0.0747 \, \text{mol NH}_4\text{Cl}}{2} \approx 0.03735 \, \text{mol} \]

### Step 6: Calculate the mass of calcium hydroxide required
Using the molar mass of Ca(OH)2:
\[ \text{Mass of Ca(OH)}_2 = \text{moles} \times \text{molar mass} = 0.03735 \, \text{mol} \times 74 \, \text{g/mol} \approx 2.77 \, \text{g} \]

### Conclusion
The weight of slaked lime required to decompose completely 4 grams of ammonium chloride is approximately **2.77 grams**.

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