What weight of potassium chlorite is required to prepare 89.6 mL of oxygen measured over water at 21.0 degrees Celsius at 743 millimeters of mercury (mmHg). The vapor pressure of water at 21.0 degrees Celsius is 18.65 mmHg
To find the weight of potassium chlorite required to prepare a certain volume of oxygen measured over water, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Total pressure (in this case, the sum of the pressure of oxygen and the vapor pressure of water)
V = Volume of gas (89.6 mL)
n = Number of moles of gas
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin
First, let's convert the temperature to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 21.0 + 273.15
T(K) = 294.15 K
Now, let's calculate the total pressure:
Total pressure = Pressure of oxygen + Vapor pressure of water
Total pressure = 743 mmHg + 18.65 mmHg
Total pressure = 761.65 mmHg
Next, convert the total pressure to atm by dividing by 760:
Total pressure = 761.65 mmHg / 760 mmHg/atm
Total pressure = 1.0028 atm
Now, rearrange the ideal gas law equation to solve for moles (n):
n = PV / RT
Substituting the values:
n = (1.0028 atm) * (89.6 mL / 1000 mL/L) / (0.0821 L·atm/mol·K) * (294.15 K)
Simplifying:
n = 0.0906 mol
Finally, we need to find the molar mass of potassium chlorite (KClO2) to calculate the weight:
Molar mass of KClO2 = (1 * molar mass of K) + (1 * molar mass of Cl) + (2 * molar mass of O)
Molar mass of KClO2 = (1 * 39.10 g/mol) + (1 * 35.45 g/mol) + (2 * 16.00 g/mol)
Molar mass of KClO2 = 122.55 g/mol
To find the weight of potassium chlorite, multiply the moles by the molar mass:
Weight = n * Molar mass
Weight = 0.0906 mol * 122.55 g/mol
Weight = 11.13 grams
Therefore, you would need approximately 11.13 grams of potassium chlorite to prepare 89.6 mL of oxygen measured over water at 21.0 degrees Celsius at 743 mmHg.