Question

5.1g of ammonia reacted with 0.2 mole of naocl according to reaction : nh3+naocl=nh2cl+naoh the formed amount of nh2cl on reaction with remaining nh3 form n2 using reaction 3nh2cl + 2nh3=n2 + 3nh4cl the amount of n2 formed (in gram) is plss answer its urgent

Answer 1

Explanation:

Given 5.1g of ammonia reacted with 0.2 mole of naocl according to reaction : nh3+naocl=nh2cl+naoh the formed amount of nh2cl on reaction with remaining nh3 form n2 using reaction 3nh2cl + 2nh3=n2 + 3nh4cl the amount of n2 formed (in gram) is plss answer its urgent

• Now moles of ammonia = mass / molecular mass
•                                      = 5.1 / 17.031
•                                      = 0.299
•                                    = 0.3
• So N h3 + NaOCl -- N h2Cl + NaOh
• So ammonia reacts with one mole of  sodium hypochlorite
• Therefore 0.2 moles of sodium hypochlorite reacts with 0.2 moles of ammonia.  
• Sodium chlorite is a limiting agent  
• so one mole of sodium chlorite produces one mole of chlorimide.
• So 0.2 moles of sodium chlorite produces 0.2 moles of Nh2CL
• Therefore remaining ammonia will be 0.3 – 0.2 = 0.1
• So the moles value need to be low.
• Now stoichiometry coefficient for Nh2Cl will be 0.2 / 3 = 0.066
• Now for ammonia, we get 0.1 / 2 = 0.05
• Since this has lowest value,it is the limiting agent.
• Therefore 2 moles of ammonia produces one mole of Nitrogen
• So 0.05 moles of ammonia produces 0.05 / 2
•                                                          = 0.025
• So mass of Nitrogen will be moles x molecular mass
•                                        = 0.025 x 28
•                                        = 0.7 g

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