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. 58.12
B. 85
C. 42.5
D. 33.3

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:
\[ \text{Ca(OH)}_2 + 2 \text{NaCl} \rightarrow \text{CaCl}_2 + 2 \text{NaOH} \]

### Step 2: Calculate the moles of NaCl used
Given the mass of NaCl is 175.5 g, we need to calculate the number of moles of NaCl. The molar mass of NaCl is calculated as follows:
- Molar mass of Na = 23 g/mol
- Molar mass of Cl = 35.5 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Now, we can calculate the moles of NaCl:
\[ \text{Moles of NaCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{175.5 \, \text{g}}{58.5 \, \text{g/mol}} \approx 3 \, \text{moles} \]

### 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, the moles of NaOH produced from 3 moles of NaCl is:
\[ \text{Moles of NaOH} = \text{Moles of NaCl} = 3 \, \text{moles} \]

### Step 4: Calculate the mass of NaOH produced theoretically
The molar mass of NaOH is calculated as follows:
- Molar mass of Na = 23 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol

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

### Step 5: Calculate the percentage yield
The actual mass of NaOH produced is given as 102 g. The percentage yield can be calculated using the formula:
\[ \text{Percentage yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage yield} = \left( \frac{102 \, \text{g}}{120 \, \text{g}} \right) \times 100 \approx 85\% \]

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

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