Answer to Question #129114 in General Chemistry for Jouaan

Question

The combustion of ammonia in the presence of oxygen yields NO2 and H2O:

4 NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)

The combustion of 43.9 g of ammonia with 258 g of oxygen produces __________ g of NO2.

Answer 1

At first, limiting reagent should be found. For this the number of moles is calculated:

n=m/M

n(NH3)=43.9/17=2.58 (mol),

n(O2)=258/32=8.06 (mol).

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 = 43.9×(4×46)/(4×17)=118.8 (g).

Answer: 118.8 g of NO2.