Answer :
The limiting reactant for the reaction of 35.8 grams of zinc, Zn and 12.9 grams of oxygen, O₂ is zinc, Zn
How to determine the limiting reactant for the reaction?
The following were obtained from the question:
- Mass of Zn before reaction = 35.8 grams
- Mass of O₂ before reaction = 12.9 grams
- Limiting reactant =?
The limiting reactant for the reaction of 35.8 grams of zinc, Zn and 12.9 grams of oxygen, O₂ is determined as follow:
2Zn + O₂ -> 2ZnO
- Molar mass of Zn = 65.38 g/mol
- Mass of Zn from the balanced equation = 2 × 65.38 = 126.76 g
- Molar mass of O₂ = 32 g/mol
- Mass of O₂ from the balanced equation = 1 × 32 = 32 g
From the balanced equation above,
126.76 g of zinc, Zn reacted with 32 g of oxygen, O₂
Therefore,
35.8 g of zinc, Zn will react with = (35.8 × 32) / 126.76 = 9.04 g of oxygen, O₂
Now, we can see from the above that only 9.04 g of oxygen, O₂ took part in the reaction.
Thus, we can conclude that the limiting reactant is zinc, Zn
Learn more about limiting reactant:
https://brainly.com/question/11587316
#SPJ1