The mass of sodium hydroxide produced when `"175.5 g of NaCl"` reacts with excess of `Ca(OH)_(2)` is 102 g. The percentage yield is :

Options

A. 0.94
B. 1.44
C. 1.5
D. 2.75

Answer

To find the percentage yield of sodium hydroxide (NaOH) produced from the reaction of sodium chloride (NaCl) with excess calcium hydroxide (Ca(OH)₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \, \text{NaCl} + \text{Ca(OH)}_2 \rightarrow 2 \, \text{NaOH} + \text{CaCl}_2 \]

### Step 2: Calculate the number of moles of NaCl
We need to calculate the number of moles of NaCl using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of NaCl = 175.5 g
- Molar mass of NaCl = 58.5 g/mol

Calculating the moles:
\[ \text{Number of moles of NaCl} = \frac{175.5 \, \text{g}}{58.5 \, \text{g/mol}} = 3 \, \text{moles} \]

### Step 3: Determine the theoretical yield of NaOH
From the balanced equation, we see that 2 moles of NaCl produce 2 moles of NaOH. Therefore, 3 moles of NaCl will produce 3 moles of NaOH.

Now, we calculate the mass of NaOH produced:
- Molar mass of NaOH = 40 g/mol
\[ \text{Mass of NaOH} = \text{Number of moles} \times \text{Molar mass} \]
\[ \text{Mass of NaOH} = 3 \, \text{moles} \times 40 \, \text{g/mol} = 120 \, \text{g} \]

### Step 4: Calculate the percentage yield
The percentage yield can be calculated using the formula:
\[ \text{Percentage yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \]
Given:
- Actual yield of NaOH = 102 g
- Theoretical yield of NaOH = 120 g

Calculating the percentage yield:
\[ \text{Percentage yield} = \left( \frac{102 \, \text{g}}{120 \, \text{g}} \right) \times 100 = 85\% \]

### Final Answer
The percentage yield of sodium hydroxide produced is **85%**.

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