The ammonia N H 3 released on quantitative reaction of 0.6 g urea N H 2 C O N H 2 with sodium hydroxide N a O H can be neutralized by

Step-by-step solution

NH 2 CONH 2   +   2 NaOH   → 2 NH 3   + Na 2 CO 3 By, stoichiometry, 1   mol of urea gives 2   mol NH 3 Moles of urea = = 0 . 6   gm 60   gm / mol = 0 . 01   mol So, 0 . 01 mole urea produce 0 . 02 mole NH 3 To neutralize ammonia HCl is required NH 3   + HCl →   NH 4 Cl By equation, it is clear that 1   mol of ammonia is neutralized by 1mol of HCl . So, 0 . 02   mol of ammonia neutralized by 0 . 02   mol of HCl ∴ Mo l e  o