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  •  The half time of first order decomposition of nitramide is 2.1 hour at <i

 The half time of first order decomposition of nitramide is 2.1 hour at \mathrm{15^{\circ}C}

If 6.2 g of  \mathrm{NH_2NO_2 } is allowed to decompose, then time taken for  \mathrm{NH_2NO_2 }  to decompose 99% and volume of dry \mathrm{ N_2O } produced at this point measured at STP will be

 

Option: 1

22.4 L 


Option: 2

22.17 L


Option: 3

2.24 L 


Option: 4

2.217 L


Answers (1)

best_answer

\mathrm{NH_2NO_2(s) \rightarrow N_2O(g) + H_2O(l)}

Initial moles of Nitramide = 6.2 g / 62 = 0.1 mol

Since, the decomposition is 99%, so 99% of the initial moles of  \mathrm{NH_2NO_2}  would be converted to \mathrm{N_2O}

Moles of  \mathrm{N_2O = 0.1 ~x~ 99/100}

volume of \mathrm{N_2O }  at STP \mathrm{= 0.1 ~x ~99~ x~ 22.4 L/ 100 = 2.217 L }.

 

Posted by

Irshad Anwar

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