Question

$2C{H}_{3}COC{H}_3+6CaOC{l}_2\to Ca{\left(C{H}_{3}CO{O}_2\right)}_2+2CHC{l}_3+3CaC{l}_2+2Ca{\left(OH\right)}_2$

If the yield of chloroform obtainable from acetone and bleaching powder is $58$%. The weight of acetone required for producing $47.8$ g of chloroform is :

• A. $40$
• B. $20$
• C. $4$
• D. $10$

Answer 1

Answer: (A)

$40$

$2C{H}_{3}COC{H}_3+6CaOC{l}_2\to Ca(C{H}_{3}CO{O}_2{)}_2+2CHC{l}_3+3CaC{l}_2+2Ca(OH{)}_2$
As yield of reaction is $58$% only, so mass of chloroform that should be produced $\frac{47.8}{\frac{58}{100}}=82.43$ g
Moles of chloroform $=\frac{82.43}{119.5}=0.69$ moles
So, moles of acetone required $0.69$ mol
Mass of acetone $0.69\times 58=40$ g