Question

50% of the reagent is used for dehydrohalogenation of 6.45 gm CH3CH2CI
What will be the weight of the main product obtained?
(At mass of H. C and Clare 1.12 & 35.5 gr/mole respectively)
(A) 14am
(B) 0.7 gm
(C) 2.8 gm
(D) 5.6 gm​

Answer 1

• Ethane is formed by dehydrohalogenation of CH3CH2Cl

CH3CH2Cl + alcoholic KOH ——› C2H5 + KCl + H2O

The molar mass of Ethane (C2H5) = (12×2 + 4×1) = 28 gm/mol

The molar mass of CH3CH2Cl = (12×2 + 1×5 + 1×35.5) = 64.5 gm/mol

Number of moles CH3CH2Cl = 6.45/64.5 = 0.1 mol

so, 0.05 mol of CH3CH2Cl will react to give 0.05 mol of Ethane

So, the mass of Ethane in 0.05 mol

= moles × molecular weight

= 0.05 mol × 28 gm/ mol

= 1.4 gm

So the answer is 1.4 gm