Oxygen gas can be generated by heating KClO3 in the presence of some MnO2.

2KClO3 → 2KCl + 3O2
If all the KClO3 in a mixture containing 1.64 g KClO3 and 0.20 g MnO2 is decomposed, how many liters of O2 gas will be generated at T = 23.0°C and P = 767.9 torr?

How many mols is 1.64 grams?

K = 39
Cl = 35.5
O3 = 3*16 = 48
so a mol is 122.5 grams
so we have 1.64/122.5 = .0134 mols KClO3

for every 2 mols KClO3 I get 3 mols O2

.0134 * 3/2 = .0201 mols O2

now at stp that is 22.4 * .0201 = .45 liters of O2
you do the PV=nRT to get volume of .0201 mols at your pressure and temperature

To find out how many liters of O2 gas will be generated, we need to use the given masses of KClO3 and MnO2, as well as the balanced equation for the reaction.

First, let's calculate the number of moles of KClO3 and MnO2:

Molar mass of KClO3 = 39.0983 g/mol (K) + 35.453 g/mol (Cl) + 3 * 16.00 g/mol (O) = 122.55 g/mol
Number of moles of KClO3 = mass of KClO3 / molar mass = 1.64 g / 122.55 g/mol

Molar mass of MnO2 = 54.938 g/mol (Mn) + 2 * 16.00 g/mol (O) = 86.94 g/mol
Number of moles of MnO2 = mass of MnO2 / molar mass = 0.20 g / 86.94 g/mol

Next, let's find the limiting reactant by comparing the number of moles of KClO3 and MnO2. The reactant that produces fewer moles of O2 will be the limiting reactant:

The balanced equation shows that 2 moles of KClO3 produce 3 moles of O2.
The molar ratio of KClO3 to O2 is 2:3.

Similarly, the balanced equation shows that 1 mole of MnO2 produces 1 mole of O2.
The molar ratio of MnO2 to O2 is 1:1.

Therefore, 2 moles of KClO3 produce 3 moles of O2, and 0.20 moles of MnO2 will produce 0.20 moles of O2.

Since 0.20 moles of MnO2 produce fewer moles of O2 compared to the moles of O2 produced by 1.64 moles of KClO3, MnO2 is the limiting reactant.

To calculate the number of moles of O2 generated, we use the molar ratio of MnO2 to O2, which is 1:1. Therefore, the number of moles of O2 produced is 0.20 moles.

Now, to find the volume of O2 gas, we can use the ideal gas law:

PV = nRT,

Where:
P = pressure = 767.9 torr
V = volume (in liters) of O2 gas (what we want to find)
n = number of moles of O2 = 0.20 moles
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin = 273.15°C + 23.0°C = 296.15 K

Rearranging the equation to solve for V, we have:

V = (n * R * T) / P

Substituting the given values:

V = (0.20 moles * 0.0821 L·atm/(mol·K) * 296.15 K) / 767.9 torr

Note: The ideal gas constant is given in units of L·atm/(mol·K), so the pressure (767.9 torr) must also be in atm for the units to cancel out correctly.

Converting torr to atm, 1 atm = 760 torr:

V = (0.20 moles * 0.0821 L·atm/(mol·K) * 296.15 K) / (767.9 torr / 760 torr/atm)

After performing the calculation, you will find the volume of O2 gas generated in liters.