Question

QUESTION 21 The combustion of ammonia in the presence of excess oxygen yields NO 2 and H 2O: 4 NH 3 (g) 7 O 2 (g) 4 NO 2 (g) 6 H 2O (g) The combustion of 14.4 g of ammonia consumes ________ g of oxygen.

Answer 1

To find the mass of oxygen consumed, we first calculate the moles of ammonia from the given mass, then use stoichiometry based on the balanced equation to find the moles of oxygen, and finally convert this to mass. The combustion of 14.4 g of ammonia consumes approximately 47.36 g of oxygen.

Step-by-step explanation:

To calculate the mass of oxygen consumed in the combustion of ammonia, we will use stoichiometry based on the balanced chemical equation provided. The balanced equation is:

4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g)

We start by calculating the number of moles of ammonia (NH3) in 14.4 g of NH3. The molar mass of NH3 is approximately 17.03 g/mol. Therefore:

14.4 g NH3 ÷ 17.03 g/mol ≈ 0.845 moles of NH3

From the balanced equation, 4 moles of NH3 require 7 moles of O2. We can calculate the moles of O2 required for 0.845 moles of NH3 using the ratio:

0.845 moles NH3 × (7 moles O2 / 4 moles NH3) ≈ 1.48 moles of O2

Finally, we convert the moles of O2 back to mass:

The molar mass of O2 is approximately 32.00 g/mol. So the mass of O2 required is:

1.48 moles O2 × 32.00 g/mol ≈ 47.36 g of O2

Therefore, the combustion of 14.4 g of ammonia consumes approximately 47.36 g of oxygen.

Answer 2

The answer to your question is 47.44 g of Oxygen

Step-by-step explanation:

Data

mass of Ammonia = 14.4 g

mass of Oxygen = ?

Balanced chemical reaction

4NH₃ + 7O₂ ⇒ 4NO₂ + 6H₂O

Process

1.- Calculate the molar mass of Ammonia

NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g

2.- Calculate the molar mass of Oxygen

O₂ = 7[16 x 2] = 7[32] = 224 g

3.- Use proportions to calculate the mass of Oxygen

68g of NH₃ --------------------- 224 g of O₂

14.4 g of NH₃ ----------------- x

x = (14.4 x 224) / 68

x = 3225.6/ 68

x = 47.44 g