UGRENT! Please help showing all work

a. Oxygen

b. 2.62 g of NO

c. 85.2%

Explanation:

Let's see the presented reaction:

4NH₃ + 5O₂  →  4NO  +  6H₂O

4 moles of ammonia react to 5 moles of oyxgen in order to produce 4 moles of NO and 6 moles of water.

To determine limting reagent, we need moles of each reactant. Let's calculate them:

3.25 g /17 g/mol = 0.191 moles

3.50 g /32 g/mol = 0.109 moles

4 moles of amonia react to 5 moles of oxygen

Then, 0.191 moles will react to (0.191 . 5) / 4 = 0.238 moles

We only have 0.109 moles available and we need 0.238, so the oxygen is the limting reactant.

5 moles of O₂ can produce 4 moles of NO

So, our 0.109 moles will produce (0.109 . 4) /5 = 0.0872 moles

We convert the moles to mass: 0.0872mol . 30g/mol = 2.62g

(Yield produced / Theoretical yield)  . 100 = %

(2.23g / 2.62g) . 100 = 85.2%

b) mutualism

hope this : )