The combustion of ammonia in the presence of excess oxygen yields NO₂and H₂O: 4 NH₃(g) + 7 O₂(g) → 4 NO₂ (g) + 6 H₂O (g)
The combustion of 43.9 g of ammonia produces ________ g of NO₂.

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of ammonia:
43.9 g NH3 x (1 mol NH3/17.03 g NH3) = 2.58 mol NH3
According to the balanced chemical equation, every 4 moles of NH3 produces 4 moles of N2O and since we have 2.58 mol NH3, we can say that:
2.58 mol NH3 x (4 mol N2O/4 mol NH3) = 2.58 mol N2O
Now we can convert the moles of N2O to grams:
2.58 mol N2O x 44.013 g N2O/mol = 113.5 g N2O
However, the question asks for the grams of N2O with one decimal place, so we round to 119 g N2O.

The Combustion of Ammonia in the Presence of Excess Oxygen