Answer to Question #170145 in General Chemistry for nouf

Question #170145

the combustion of 43.9 g ammonia in the presence of oxygen yields no and h2o


1
Expert's answer
2021-03-12T05:59:27-0500

Solution:


Initially, limiting reagent should be found:

"n=\\frac{m}{M}"

n(NH3)="\\frac{43.9}{17}=2.58 moles"


n(O2)="\\frac{258}{32}=8.06moles"


Mole ratio is NH3:O2=1:3.12=4:12.5.


Ratio needed for the reaction is 4:7, so NH3 is a limiting reagent, and it's mass is used for further calculations.

Mass of NO2 may be found from a proportion:

x = "\\dfrac{43.9(4*46)}{4*17}=118.8g"


Answer: 118.8 g of NO2.


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