Question

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

Answer 1

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.