If the yield of chloroform obtained from acetone and bleaching powder and bleaching powder is $58 %$ . The weight of acetone required for producing $47.847.8 g$ of chloroform is:
$2 C H_{3} C O C H_{3} + 6 C a O C l_{2} \to C a (C H_{3} C O O)_{2} + 2 C H C_{3} + 3 C a C l_{2} + 2 C a (O H)_{2}$ $

Open in App

Solution

2 moles of Acetone gives 2 moles of $C H C l_{3}$ .
or 1 mole of $C H_{3} C O C H_{3}$ gives 1 mole of $C H C l_{3}$ .
Given mole of $C H C l_{3} = \frac{47.847}{119.5} = 0.4$
moles of $C H_{3} C O C H_{3} = 0.4$
Weight of $C H_{3} C O C H_{3} = 0.4 \times 58 = 23.2 g$
But the yield is $58 %$ , Hence weight of $C H_{3} C O C H_{3}$ required $= \frac{23.2}{58} \times 100 = 40 g$

Suggest Corrections

Similar questions

2 C H_{3} C O C H_{3} + 6 C a O C l_{2} \to C a {(C H_{3} C O O_{2})}_{2} + 2 C H C l_{3} + 3 C a C l_{2} + 2 C a {(O H)}_{2}

58

47.8

Join BYJU'S Learning Program