Potassium perchlorate decomposes as follows: KClO4 (s) --> KCl (s) + 2 O2 (g)
If 269 mL of wet O2 (g) is collected over water at 25 c (vapor pressure= 23.8 torr) and the barometric pressure is 748.0 torr, how many grams of potassium perchlorate have decomposed?
Use Ideal Gas Law (PV=nRT & n=PV/RT)=> moles of O2... Divide by 2 => moles of KClO3 ... Multiply moles KClO3 by formula wt of KClO3 => grams of KClO3 decomposed. Be sure all data is converted to same units as R-value. I get 0.021 gms KClO3. Good luck.
Correction... I'm used to decomposing KClO3, should of used f.Wt. of KClO4 = 138g/mole... Same logic applies to all else. Moles KClO4 decomposed = 0.0017mole O2, then => I get 0.235g KClO4 decomposed ... Sorry bout that.
correction-2 => 0.0017 mole KClO4! not O2.
To find the number of grams of potassium perchlorate that have decomposed, we need to use the ideal gas law equation to determine the moles of oxygen gas, and then use stoichiometry to relate it to the moles of potassium perchlorate.
Let's break down the steps:
Step 1: Convert the volume of wet O2 gas to the volume of dry O2 gas.
The collected volume of wet O2 gas needs to be corrected for the vapor pressure of water at 25 °C. The partial pressure of water vapor is 23.8 torr. So, we subtract this value from the total pressure to obtain the partial pressure of dry O2 gas.
Partial pressure of O2 gas = Total pressure - Vapor pressure of water
Partial pressure of O2 gas = 748.0 torr - 23.8 torr = 724.2 torr
Step 2: Convert torr to atm.
We need to convert the pressure from torr to atm. To do this, we divide the pressure in torr by 760 torr/atm.
Partial pressure of O2 gas (in atm) = 724.2 torr / 760 torr/atm = 0.952 atm
Step 3: Convert the volume of O2 gas to moles.
We can use the ideal gas law equation to convert the volume of dry O2 gas to moles.
PV = nRT
n = PV / RT
Since we are given the pressure, volume, temperature, and the ideal gas constant (R = 0.0821 L·atm/(mol·K)), we can substitute those values into the equation.
n (moles of O2) = (0.952 atm) × (0.269 L) / (0.0821 L·atm/(mol·K)) × (298 K)
n (moles of O2) ≈ 0.0112 moles
Step 4: Use the stoichiometry of the balanced equation to calculate the moles of potassium perchlorate.
From the balanced equation, we can see that 1 mole of KClO4 decomposes to produce 2 moles of O2 gas.
So, 0.0112 moles of O2 gas would correspond to 0.0056 moles of KClO4 decomposed.
Step 5: Convert moles of KClO4 to grams.
To convert moles of KClO4 to grams, we need to multiply by the molar mass of KClO4, which is calculated by adding up the atomic masses of potassium (K), chlorine (Cl), and four oxygen (O) atoms.
Molar mass of KClO4 = 39.1 g/mol + 35.5 g/mol + (4 × 16.0 g/mol) = 122.6 g/mol
Mass of KClO4 decomposed = 0.0056 moles × 122.6 g/mol
Mass of KClO4 decomposed ≈ 0.6872 grams
Therefore, approximately 0.6872 grams of potassium perchlorate have decomposed.