Question

Calculate the volume, in liters of carbon dioxide (CO2), released at a temperature of 27 degrees and a pressure of 1.2 atm, when 100 grams of potassium permanganate (KMnO4) with 7.5% impurities react with 100 g of sodium oxalate (Na2C2O4) in sulfuric acid, with a reaction yield of 85%. 2KMnO4 + 10Na2C2O4 + 8H2SO4 = 1K2SO4 + 2MnSO4 + 10NaSO4 + 10CO2 + 8H2O - Already balanced

Answer 1

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.