Calculate the weight of potassium chlorate required to prepare 380 mL of wet oxygen collected over water at 20 degrees C and 750 mm of Hg.
Use PV = nRT to convert 380 mL of wet oxygen to dry oxygen at STP. Then use this example.
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Why did the oxygen go to therapy?
Because it needed some space to breathe! But let's break down this question, shall we?
To calculate the weight of potassium chlorate required, we first need to determine the volume of dry oxygen. Since the oxygen is collected over water, we have to take into account the vapor pressure of water at 20 degrees C and 750 mm Hg.
But remember, I'm a Clown Bot, not a chemistry bot! So I would need some values to assist you further.
To calculate the weight of potassium chlorate required to prepare 380 mL of wet oxygen collected over water at 20 degrees C and 750 mm of Hg, we need to follow the following steps:
Step 1: Determine the total pressure of the wet oxygen:
The total pressure can be calculated by adding the partial pressure of water vapor (from the vapor pressure table at the given temperature) to the given pressure of 750 mmHg.
Step 2: Calculate the moles of oxygen gas:
To calculate the moles of oxygen gas, we use the ideal gas law equation: PV = nRT, where P is the total pressure of the wet oxygen, V is the volume of the wet oxygen, n is the number of moles of oxygen gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Step 3: Convert moles of oxygen to moles of potassium chlorate:
The balanced equation for the decomposition of potassium chlorate is:
2KClO3 → 2KCl + 3O2
From the balanced equation, we can see that for every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are consumed.
Step 4: Calculate the weight of potassium chlorate:
To calculate the weight of potassium chlorate, we'll multiply the moles of potassium chlorate from step 3 by its molar mass.
Now let's perform the calculations:
Step 1: Determine the total pressure of the wet oxygen:
Total pressure = 750 mmHg + vapor pressure of water at 20 degrees C (from vapor pressure table) = X mmHg
Step 2: Calculate the moles of oxygen:
PV = nRT
n = (X mmHg * 380 mL) / (R * (20 + 273.15) K)
Step 3: Convert moles of oxygen to moles of potassium chlorate:
For every 3 moles of oxygen, 2 moles of potassium chlorate are consumed.
Step 4: Calculate the weight of potassium chlorate:
Weight of potassium chlorate = (moles of potassium chlorate) * (molar mass of potassium chlorate)
Please note that the molar mass of potassium chlorate is KClO3 = 39.0983 g/mol (Potassium) + 35.453 g/mol (Chlorine) + 3 * 15.999 g/mol (Oxygen).
To calculate the weight of potassium chlorate required to prepare wet oxygen, we need to understand a few concepts.
First, when oxygen is collected over water, it becomes moist or wet due to the presence of water vapor. We need to consider the partial pressure of water vapor in the final calculation.
Secondly, we need to consider the temperature and pressure at which the oxygen is collected. This helps us determine the molar volume of the gas.
Now, let's break down the process step by step:
Step 1: Calculate the partial pressure of water vapor.
At 20 degrees C and 750 mm Hg, we can use a water vapor pressure table or equation to find the partial pressure of water vapor. The partial pressure of water vapor at 20 degrees C is approximately 17.5 mm Hg.
Step 2: Calculate the partial pressure of oxygen.
To find the partial pressure of oxygen, we need to subtract the partial pressure of water vapor from the total pressure.
Partial pressure of oxygen = Total pressure - Partial pressure of water vapor
Partial pressure of oxygen = 750 mm Hg - 17.5 mm Hg = 732.5 mm Hg
Step 3: Calculate the molar volume of the gas.
The molar volume of a gas at STP (Standard Temperature and Pressure) is 22.4 L/mol. However, since we are not at STP, we need to adjust the molar volume based on the temperature and pressure.
Molar volume = (Volume of gas collected) * (Total Pressure) / (Partial Pressure of oxygen)
Molar volume = 380 mL * (750 mm Hg / 732.5 mm Hg) = 388.14 mL
Step 4: Calculate the moles of oxygen gas.
Moles of oxygen gas = (Molar volume) / (Molar volume at STP)
Moles of oxygen gas = 388.14 mL / 22.4 L/mol = 0.01732 moles
Step 5: Calculate the molar mass of potassium chlorate (KClO3).
The molar mass of potassium chlorate is K (39.1 g/mol) + Cl (35.45 g/mol) + 3 O (16.00 g/mol) = 122.55 g/mol.
Step 6: Calculate the weight of potassium chlorate required.
Weight of potassium chlorate = (Moles of oxygen gas) * (Molar mass of KClO3)
Weight of potassium chlorate = 0.01732 moles * 122.55 g/mol = 2.13 grams
Therefore, the weight of potassium chlorate required to prepare 380 mL of wet oxygen collected over water at 20 degrees C and 750 mm Hg is approximately 2.13 grams.