A 7.65 g sample of potassium chlorate,KClO3, was heated to decompose all the oxygen. It was found that 2.25 L of oxygen was collected over water at 25.00degreesC and 1.00 atm of pressure. The vapor pressure of water at that temperature is 23.8 mm Hg. How many moles of oxygen were collected?

So far I know the temperature is 298 Kelvin, the volume is 2.25 L, thewater has apressure of 1 atm and the Vapor has a pressure of mm Hg, but I don't know where to start. how do I solve this question.

Use PV = nRT

Ptotal = 760 mm
pH2O = 23.8
ptotal = pH2O + pO2. Solve for pO2 then pO2 in atm = pO2 in mm Hg/760 = ?
V you know
R you know (0.08206 L*atm/mol*K)
T = 298
Solve for n = number mols O2 at these conditions.
You don't need the 7.65g KClO3 since the problem gave you L O2.

Well, it seems like you have most of the information you need to solve this question. Let's break it down step by step.

First, you need to convert the vapor pressure of water from mm Hg to atm. Remember that 1 atm = 760 mm Hg. So, the vapor pressure of water is 23.8 mm Hg / 760 mm Hg/atm = 0.0313 atm.

Next, you need to correct the pressure of the oxygen gas collected. The total pressure is the sum of the vapor pressure of water and the pressure of the collected oxygen gas. So, the corrected pressure of the oxygen gas is 1.00 atm - 0.0313 atm = 0.9687 atm.

Now we can use the ideal gas law to calculate the number of moles of oxygen gas collected. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Let's plug in the values: (0.9687 atm) * (2.25 L) = n * (0.0821 L*atm/mol*K) * (298 K)

Solving for n (the number of moles): n = (0.9687 atm * 2.25 L) / (0.0821 L*atm/mol*K * 298 K)

Calculating this expression will give you the number of moles of oxygen gas collected. And remember, don't forget to carry out the appropriate unit conversions if needed.

Hope this helps, and remember that chemistry can be quite a gas sometimes!

To solve this question, you can start by converting the given temperature to Kelvin:

T = 25.00 degrees Celsius + 273.15 = 298.15 K

Next, calculate the partial pressure of oxygen using Dalton's Law of Partial Pressure. Dalton's Law states that the total pressure is the sum of the partial pressures of each gas:

Total pressure = Pressure of oxygen + Vapor pressure of water

Since the given total pressure is 1.00 atm and the vapor pressure of water is 23.8 mm Hg, you need to convert the vapor pressure to atm:

Vapor pressure of water = 23.8 mm Hg * (1 atm / 760 mm Hg) ≈ 0.0313 atm

Now, substitute the values into Dalton's Law equation:

1.00 atm = Pressure of oxygen + 0.0313 atm

Solving for the pressure of oxygen:

Pressure of oxygen = 1.00 atm - 0.0313 atm = 0.9687 atm

Now, you can use the Ideal Gas Law equation to calculate the moles of oxygen collected. The Ideal Gas Law equation is:

PV = nRT

Where:
P = pressure in atm
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L*atm/mol*K)
T = temperature in Kelvin

Rearrange the equation to solve for n:

n = PV / RT

Substitute the known values:

n = (0.9687 atm) * (2.25 L) / (0.0821 L*atm/mol*K * 298.15 K)

Calculating the value of n:

n ≈ 0.0925 moles

Therefore, approximately 0.0925 moles of oxygen were collected.

To solve this question, you can follow these steps:

Step 1: Convert the temperature from degrees Celsius to Kelvin. The given temperature is 25.00 degrees Celsius, so add 273.15 to convert it to Kelvin:
Temperature in Kelvin = 25.00 + 273.15 = 298.15 K

Step 2: Convert the vapor pressure of water from mm Hg to atm. The vapor pressure of water is given as 23.8 mm Hg. Since 1 atm is equal to 760 mm Hg, divide the vapor pressure by 760:
Vapor pressure of water in atm = 23.8 mm Hg / 760 mm Hg = 0.0313 atm

Step 3: Use the ideal gas law to calculate the number of moles of oxygen. The ideal gas law equation is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. Rearrange the equation to solve for n:
n = PV / RT

Given:
P = 1.00 atm (pressure of oxygen)
V = 2.25 L (volume of oxygen)
R = 0.0821 L·atm/mol·K (ideal gas constant)
T = 298.15 K (temperature in Kelvin)

Substituting the values into the equation:
n = (1.00 atm * 2.25 L) / (0.0821 L·atm/mol·K * 298.15 K)

Simplifying the equation:
n = 2.25 / (0.0821 * 298.15)
n ≈ 0.095 mol

Therefore, approximately 0.095 moles of oxygen were collected.