Question

5.1 g 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 is


1.4 g


0.5 g


1.95 g


2.24 g

Answer 1

$\begin{lgathered}\red\longmapsto\:\rm\tt\green{x+20=5(x-20)}\\ \\ \red\longmapsto\:\rm\tt\blue{x+20=5x-100}\\ \\ \red\longmapsto\:\rm\tt\orange{x-5x=-100-20}\\ \\ \red\longmapsto\:\rm\tt\pink{-4x=-120}\\ \\ \red\longmapsto\:\rm\tt\purple{x=\cancel\dfrac{-120}{-4}=30}\\ \\ \red\longmapsto\:\large\underline{\boxed{\bf\tt\green{x}\tt\orange{=}\tt\red{30}}}\end{lgathered}$

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$\tt{Correct\: option =24g}$