Question

If the yield of chloroform obtainable from acetone and bleaching powder is 58%. What is the weight of acetone (in gm) required for producing 239 mg of chloroform ?

$2CH_3COCH_3 + 6CaOCl_2 \rightarrow Ca(CH_3COO)_2 + 2CHCl_3 + 3CaCl_2 + 2Ca(OH)_2$

Answer 1

0.2

Answer 2

The problem involves calculating the amount of a reactant needed to produce a specific amount of product, considering the reaction's yield.

1. Write down the balanced chemical equation: The given reaction is: $2CH_3COCH_3 + 6CaOCl_2 \rightarrow Ca(CH_3COO)_2 + 2CHCl_3 + 3CaCl_2 + 2Ca(OH)_2$

2. Determine the molar masses of the relevant compounds:

• Acetone ($CH_3COCH_3$): Molar mass = $(3 \times 12.01) + (6 \times 1.008) + (1 \times 16.00) = 36.03 + 6.048 + 16.00 = 58.078 \text{ g/mol} \approx 58 \text{ g/mol}$
• Chloroform ($CHCl_3$): Molar mass = $(1 \times 12.01) + (1 \times 1.008) + (3 \times 35.45) = 12.01 + 1.008 + 106.35 = 119.368 \text{ g/mol} \approx 119.5 \text{ g/mol}$ (using 35.5 for Cl as is common in such problems)

3. Determine the mole ratio from the balanced equation: From the equation, 2 moles of acetone ($CH_3COCH_3$) produce 2 moles of chloroform ($CHCl_3$). This simplifies to a 1:1 mole ratio: 1 mole of acetone produces 1 mole of chloroform.

4. Calculate the moles of chloroform required: Given mass of chloroform = 239 mg = 0.239 g Moles of $CHCl_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.239 \text{ g}}{119.5 \text{ g/mol}} = 0.002 \text{ mol}$

5. Calculate the theoretical moles of acetone required: Since the mole ratio of acetone to chloroform is 1:1, the theoretical moles of acetone required are equal to the moles of chloroform produced. Theoretical moles of $CH_3COCH_3 = 0.002 \text{ mol}$

6. Calculate the theoretical mass of acetone required: Theoretical mass of $CH_3COCH_3 = \text{Theoretical moles} \times \text{Molar mass}$ Theoretical mass of $CH_3COCH_3 = 0.002 \text{ mol} \times 58 \text{ g/mol} = 0.116 \text{ g}$

7. Account for the reaction yield: The yield of chloroform is 58%. This means that to obtain 0.239 g of chloroform, we need to start with more acetone than the theoretical amount calculated (which assumes 100% yield). The formula to find the actual reactant needed is: Actual mass of reactant = $\frac{\text{Theoretical mass of reactant}}{\text{Yield percentage}/100}$ Actual mass of acetone = $\frac{0.116 \text{ g}}{0.58}$ Actual mass of acetone = $0.2 \text{ g}$

The weight of acetone required is 0.2 gm.