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%

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.