Answer to Question #253718 in General Chemistry for Bang-bang

Question

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

4NH3(g)--4NO2(g)6H2O(g)

Determine the mass in grams of NO2 when 43.9g of NH3 was used.

Answer 1

$4NH_3\to 4NO_2+6H_2O$

moles of ammonia$=\frac{mass}{RMM}=\frac{43.9}{17}=2.582$

mole ration $NH_3:NO_2=1:1$

moles of $NO_2=2.582$

Mass of $NO_2=number \space of \space moles \times RMM=2.582\times (14+16+16)=118.772g$