The combustion of ammonia in the presence of excess oxygen yields NO₂and H₂O:

The combustion of 28.8 g of ammonia consumes g of oxygen.

Question

The combustion of ammonia in the presence of excess oxygen yields NO₂and H₂O: The combustion of 28.8 g of ammonia consumes g of oxygen. A) 108 B) 94.9 C) 28.8 D) 54.1 E) 15.3

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for the combustion of ammonia:

4NH3 + 7O2 → 4NO + 6H2O

This tells us that 4 moles of NH3 react with 7 moles of O2 to produce 4 moles of NO and 6 moles of H2O.

Next, we need to find the number of moles of NH3 in 28.8 g:

28.8 g NH3 × (1 mol NH3/17.03 g NH3) = 1.6910 mol NH3

Since ammonia is present in excess, we know that all 1.6910 mol of NH3 will react with oxygen to form NO and H2O.

Using the mole ratio from the balanced equation, we can calculate the number of moles of O2 required:

7 mol O2/4 mol NH3 × 1.6910 mol NH3 = 2.949 mol O2

Finally, we can convert this to grams of O2:

2.949 mol O2 × 32.00 g O2/mol = 94.4 g O2

Therefore, the answer is B) 94.9 (rounded to two significant figures to match the given mass of NH3).

The Combustion of Ammonia in the Presence of Excess Oxygen