Answer:
a. Oxygen
b. 2.62 g of NO
c. 85.2%
Step-by-step 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%