Answer:
Explanation:
<u>1. Molecular chemical equation:</u>
- 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
<u>2. Mole ratios:</u>
- 2 mol KClO₃ : 2 mol KCl : 3 mol O₂
<u>3. Number of moles of KClO₃</u>
- Number of moles = mass in grams / molar mass
- Molar mass of KClO₃ = 122.55 g/mol
- Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol
<u>3. Number of moles of O₂</u>
As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.
- 3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃
- x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂
Round to 3 significant figures: 0.665 mol of O₂ ← answer
The answer to your question big man would be E
Answer:
The rate at which 
is being produced is 0.0228 M/s.
The rate at which 
is being consumed is 0.0912 M/s.
Explanation:
Rate of the reaction : R
![R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B-1%7D%7B4%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
The rate at which hydrogen is being formed = ![\frac{d[H_2]}{dt}=0.137 M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D0.137%20M%2Fs)
![R=\frac{1}{6}\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
The rate at which is being produced:
![R=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
![0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=0.0228%20M%2Fs%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
The rate at which is being consumed :
![R=\frac{-1}{4}\frac{d[PH_3]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B-1%7D%7B4%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D)
![0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}](https://tex.z-dn.net/?f=0.0228%20M%2Fs%5Ctimes%204%3D%5Cfrac%7B-1%7D%7B1%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D)
![\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B1%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D%3D0.912%20M%2Fs)
