One way to generate oxygen is to heat potassium chlorate, KCLO3 (the other product is potassium chloride). If 386 mL of oxygen at 41 degree centigrade and 97.8 kPa is generated by this reaction, what is the minimum mass of KCLO3 used?

2KClO3 ==> 2KCl + 3O2

Use PV = nRT and calculate n for Oxygen.
Using the coefficients in the balanced equation, convert mols O2 to mols KClO3.
Now convert mols KClO3 to grams. g = mols x molar mass = ?

To find the minimum mass of KCLO3 used, we need to use the Ideal Gas Law equation:

PV = nRT

where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (8.31 J/(mol K))
T = temperature (in Kelvin)

First, we need to convert the given values to the appropriate units:
- The temperature needs to be converted from degrees Celsius to Kelvin: T = 41°C + 273.15 = 314.15 K.
- The pressure is already in the correct unit (kPa).

Next, we need to calculate the number of moles of oxygen using the Ideal Gas Law equation. Rearranging the equation, we have:

n = PV / RT

Substituting the given values:
n = (97.8 kPa * 0.386 L) / (8.31 J/(mol K) * 314.15 K)

Now, let's convert liters to moles and simplify:
n = (97.8 kPa * 0.386 L) / (8.31 J/(mol K) * 314.15 K)
n = (97.8 * 0.386) / (8.31 * 314.15)
n = 0.0332372 moles of oxygen

According to the balanced chemical equation, the ratio between KCLO3 and oxygen is 2:3. This means that for every 2 moles of KCLO3, 3 moles of oxygen are produced.

Using this ratio, we can calculate the moles of KCLO3:
moles of KCLO3 = (2/3) * moles of oxygen
moles of KCLO3 = (2/3) * 0.0332372
moles of KCLO3 = 0.0221581 moles of KCLO3

Now, we can calculate the molar mass of KCLO3, which is the mass of 1 mole:
Molar mass of KCLO3 = mass of KCLO3 / moles of KCLO3

Assuming the molar mass of potassium chlorate (KCLO3) is 122.55 g/mol, we can rearrange the formula to solve for mass:
Mass of KCLO3 = molar mass of KCLO3 * moles of KCLO3
Mass of KCLO3 = 122.55 g/mol * 0.0221581 moles

Calculating the final result:
Mass of KCLO3 = 2.71318 grams

Therefore, the minimum mass of KCLO3 used is approximately 2.71318 grams.

To find the minimum mass of KCLO3 used, we can use the ideal gas law and the molar volume of a gas at STP.

First, we need to convert the given temperature from degrees Celsius to Kelvin. We can do this by adding 273.15 to the temperature:
Temperature in Kelvin = 41°C + 273.15 = 314.15 K

Next, we can use the ideal gas law equation:
PV = nRT

Where:
P = pressure (97.8 kPa)
V = volume (386 mL or 0.386 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (314.15 K)

We can rearrange the equation to solve for the number of moles (n):
n = PV / RT

Substituting the given values:
n = (97.8 kPa)(0.386 L) / (0.0821 L·atm/mol·K)(314.15 K)

Now, we need to convert the pressure from kilopascals (kPa) to atmospheres (atm) since the ideal gas constant has units for atm:
n = (97.8 kPa)(0.386 L) / (0.0821 L·atm/mol·K)(314.15 K) * (1 atm / 101.325 kPa)

Calculating this expression:
n ≈ 0.01457 moles

Since the reactant potassium chlorate (KCLO3) produces 1 mole of oxygen (O2) gas, the molar ratio is 1:1. This means that 0.01457 moles of KCLO3 will also produce 0.01457 moles of oxygen gas.

To find the molar mass of KCLO3, we use the atomic masses of potassium (K), chlorine (Cl), and oxygen (O):
K = 39.1 g/mol
Cl = 35.5 g/mol
O = 16 g/mol

Molar mass of KCLO3 = potassium (K) + chlorine (Cl) + 3 × oxygen (O)
Molar mass of KCLO3 = 39.1 g/mol + 35.5 g/mol + 3 × 16 g/mol
Molar mass of KCLO3 = 122.6 g/mol

Finally, we can calculate the minimum mass of KCLO3 used:
Mass = moles × molar mass
Mass = 0.01457 moles × 122.6 g/mol

Calculating this expression:
Mass ≈ 1.78 grams

Therefore, the minimum mass of KCLO3 used to generate the given amount of oxygen is approximately 1.78 grams.