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
To solve this problem, we first need to calculate the moles of KMnO4 and Na2C2O4 used in the reaction.
1. Moles of KMnO4:
Molar mass of KMnO4 = 39.0983 g/mol (K) + 54.938045 g/mol (Mn) + 4 * 15.999 g/mol (O) = 158.0339 g/mol
Moles of KMnO4 = mass of KMnO4 / molar mass of KMnO4
= 100 g / 158.0339 g/mol
= 0.632 mol
2. Moles of Na2C2O4:
Molar mass of Na2C2O4 = 2 * 22.989769 g/mol (Na) + 2 * 12.0107 g/mol (C) + 4 * 15.999 g/mol (O) = 134.00 g/mol
Moles of Na2C2O4 = mass of Na2C2O4 / molar mass of Na2C2O4
= 100 g / 134.00 g/mol
= 0.746 mol
Next, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction. We do this by comparing the moles of each reactant used.
From the balanced equation, we can see that the stoichiometric ratio between KMnO4 and Na2C2O4 is 2:10. Therefore, we need 10/2 = 5 moles of Na2C2O4 for every mole of KMnO4.
Since we have 0.632 moles of KMnO4, the equivalent moles of Na2C2O4 required would be:
0.632 mol KMnO4 * 5 mol Na2C2O4 / 2 mol KMnO4 = 1.58 mol Na2C2O4
Comparing this with the actual moles of Na2C2O4 (0.746 mol), we can see that Na2C2O4 is the limiting reactant.
Now, we can calculate the moles of CO2 produced based on the balanced equation.
From the balanced equation, the stoichiometric ratio between Na2C2O4 and CO2 is 10:10. Therefore, 10 moles of CO2 are produced for every 10 moles of Na2C2O4.
Moles of CO2 produced = moles of Na2C2O4 * (10 mol CO2 / 10 mol Na2C2O4)
= 0.746 mol * (10 mol CO2 / 10 mol Na2C2O4)
= 0.746 mol CO2
Now, we can calculate the volume of CO2 at the given temperature and pressure.
Ideal Gas Equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters
n = moles
R = gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin
First, we need to convert the temperature from degrees Celsius to Kelvin.
T(K) = T(°C) + 273.15
= 27 + 273.15
= 300.15 K
Now, we can calculate the volume of CO2 using the ideal gas equation.
V = (n * R * T) / P
= (0.746 mol * 0.0821 L.atm/mol.K * 300.15 K) / 1.2 atm
≈ 15.81 L
Therefore, the volume of carbon dioxide released is approximately 15.81 liters.
To determine the volume of carbon dioxide (CO2) released, we need to use the Ideal Gas Law equation, which is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's calculate the number of moles of CO2 produced. We'll need the molar mass of CO2, which is 44.01 g/mol.
Given that the reaction yield is 85%, we can calculate the moles of CO2 produced.
Moles of CO2 = (mass of CO2 produced * reaction yield) / molar mass of CO2
Mass of CO2 produced:
The balanced equation shows that the ratio of KMnO4 to CO2 is 2:10. Thus, for every 2 moles of KMnO4, 10 moles of CO2 are produced.
Molar mass of KMnO4 = (39.10 g/mol * 1) + (54.938 g/mol * 1) + (16.00 g/mol * 4)
= 39.10 + 54.938 + 64.00
= 158.038 g/mol
Moles of KMnO4 = mass of KMnO4 / molar mass of KMnO4 = 100 g / 158.038 g/mol
≈ 0.633 mol
Moles of CO2 = (0.633 mol KMnO4 * 10 mol CO2) / 2 mol KMnO4
= 3.165 mol
Next, let's convert the temperature from degrees Celsius to Kelvin:
T(in Kelvin) = T(in Celsius) + 273.15
T = 27 + 273.15
T = 300.15 K
Now, we can calculate the volume of CO2 using the Ideal Gas Law equation:
PV = nRT
V = (nRT) / P
= (3.165 mol * 0.0821 L·atm/(mol·K) * 300.15 K) / 1.2 atm
≈ 63.04 L
Therefore, the volume of carbon dioxide (CO2) released is approximately 63.04 liters.