Question

Complete and balance the following equations:
(a) (C_2H_4Br_2 + 2KOH (textalcoholic) )
(b) (8NH_3(textexcess) + 3Cl_2 )
(c) (C_12H_22O_11 xrightarrowtextconc. H_2SO_4)

Hint:

In equation (b), if Chlorine were in excess instead of Ammonia, the product would be the highly explosive liquid, Nitrogen Trichloride (NCl_3)!

Answer 1

(a) For C₂H₄Br₂ + 2KOH (alcoholic):

Alcoholic KOH causes elimination (dehydrohalogenation). Two molecules of HBr are removed forming an alkyne.

Balanced equation:

C₂H₄Br₂ + 2KOH (alcoholic) → C₂H₂ + 2KBr + 2H₂O

(b) For 8NH₃ (excess) + 3Cl₂:

With excess ammonia, nitrogen gas and ammonium chloride are formed.

Balanced equation:

8NH₃ + 3Cl₂ → 6NH₄Cl + N₂

(c) For C₁₂H₂₂O₁₁ (conc. H₂SO₄):

Concentrated sulphuric acid acts as a dehydrating agent and removes water from sugar.

Balanced equation:

C₁₂H₂₂O₁₁ → 12C + 11H₂O

Final Answers:
(a) C₂H₄Br₂ + 2KOH → C₂H₂ + 2KBr + 2H₂O
(b) 8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
(c) C₁₂H₂₂O₁₁ → 12C + 11H₂O