If the yield of chloroform obtainable from acetone and bleaching powder is 58% . What is the weight of acetone required for producing 47.8 gm of chloroform ?
`2CH_(3)COCH_(3) +6CaOCl_(2) to Ca(CH_(3)C"OO")_(2)`
`+ 2CHCl_(3) + CaCl_(2) + 2Ca(OH)_(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the mass of chloroform that should be produced.
Given that the yield of chloroform is 58%, we can find the theoretical mass of chloroform that should be produced using the formula:

\[ \text{Mass of chloroform (theoretical)} = \frac{\text{Actual mass of chloroform}}{\text{Yield}} \]

Substituting the values:

\[ \text{Mass of chloroform (theoretical)} = \frac{47.8 \, \text{g}}{0.58} = 82.41 \, \text{g} \]

### Step 2: Calculate the moles of chloroform produced.
Next, we need to calculate the number of moles of chloroform produced. The molar mass of chloroform (CHCl₃) is approximately 119.5 g/mol. We can use the formula:

\[ \text{Moles of chloroform} = \frac{\text{Mass of chloroform}}{\text{Molar mass of chloroform}} \]

Substituting the values:

\[ \text{Moles of chloroform} = \frac{82.41 \, \text{g}}{119.5 \, \text{g/mol}} \approx 0.69 \, \text{mol} \]

### Step 3: Determine the moles of acetone required.
From the balanced chemical equation, we see that 2 moles of acetone (CH₃COCH₃) produce 2 moles of chloroform (CHCl₃). Therefore, the moles of acetone required are equal to the moles of chloroform produced:

\[ \text{Moles of acetone} = 0.69 \, \text{mol} \]

### Step 4: Calculate the mass of acetone required.
The molar mass of acetone (CH₃COCH₃) is approximately 58 g/mol. We can find the mass of acetone required using the formula:

\[ \text{Mass of acetone} = \text{Moles of acetone} \times \text{Molar mass of acetone} \]

Substituting the values:

\[ \text{Mass of acetone} = 0.69 \, \text{mol} \times 58 \, \text{g/mol} \approx 40.02 \, \text{g} \]

### Final Answer:
The weight of acetone required for producing 47.8 g of chloroform is approximately **40.02 g**.

---