Answer:
V of ( CO2) ≈ 2.07
Step-by-step explanation:
let's break down the steps for calculating the volume of carbon dioxide (CO2) released:
Step 1: Calculate moles of CO2 produced.
Given:
Mass of KMnO4 = 100 grams
Molar mass of KMnO4 = 158.034 g/mol
calculate moles of KMnO4:
n(KMnO4) = Mass/ Molar mass
n(KMnO4) = 100 g / 158.034 g/mol
≈ 0.632mol
Using the balanced chemical equation, you know that for every 2 moles of KMnO4, 10 n(CO2) are produced.
So, calculate n(CO2) produced:
n(CO2) produced = (2 n(KMnO4) / 10 n(CO2) * n(KMnO4)
But n(KMnO4) = 0.632 mol
n(CO2) produced ≈ (2/10) * 0.632
≈ 0.1264 mol
Step 2: Calculate the volume of CO2 using the ideal gas law.
Given:
Pressure (P) = 1.2 atm
Ideal gas constant (R) = 0.0821 L.atm/mol.K
Temperature (T) = 27°C = (373 + 27) = 300.15 K
Use the ideal gas law formula:
V = (nRT) / P
Now, inserting the values:
V = (0.1264 moles * 0.0821 L.atm/mol.K * 300.15 K) / 1.2 atm
Step 3: Calculate the volume.
Now, calculate the volume of CO2:
V ≈ (0.1264 * 0.0821 * 300.15) / 1.2 ≈ 2.07 liters
So, approximately 2.07 liters of CO2 are released in this reaction.