When a sample of potassium nitrate is heated, oxygen gas is produced. This gas, collected in a 750 milliliter flask, has a pressure of 2.8 atmosphere and the temperature is recorded to be 53.6°C. How many moles of oxygen gas are produced in this reaction?

0.078 mol.

To calculate the number of moles of oxygen gas produced, we can use the Ideal Gas Law equation:

PV = nRT

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

First, let's convert the temperature from Celsius to Kelvin:
T = 53.6°C + 273.15 = 326.75 K

Now, let's convert the volume from milliliters to liters:
V = 750 mL = 0.75 L

Next, let's plug the values into the formula:

2.8 atm * 0.75 L = n * 0.0821 L·atm/mol·K * 326.75 K

Simplifying:

2.1 L·atm = n * 26.81 L·atm/mol

Now, solve for n:

n = 2.1 L·atm / 26.81 L·atm/mol

n ≈ 0.0783 mol

Therefore, approximately 0.0783 moles of oxygen gas are produced in this reaction.

To calculate the number of moles of oxygen gas produced, we can use the Ideal Gas Law formula:

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)

Given data:
P = 2.8 atm
V = 750 mL = 0.75 L
T = 53.6°C = 53.6 + 273.15 = 326.75 K

Let's plug in the values into the formula:

2.8 atm * 0.75 L = n * 0.0821 L.atm/mol.K * 326.75 K

Simplifying the equation:

2.1 L.atm = 26.897775 L.atm/mol

Dividing both sides by 26.897775 L.atm:

2.1 L.atm / 26.897775 L.atm/mol = n

n ≈ 0.0781 mol

Therefore, approximately 0.0781 moles of oxygen gas are produced in this reaction.