can anyone help with this problem please.....

Potassium chlorate decomposes according to the following equation:
2KClO3 (s) → 2KCl (s) + 3O2 (g).
If a 3.00 g sample of KClO3 is decomposed and the oxygen is collected at 24.0 °C and 0.991 atm, what volume of oxygen gas will be collected? Use the ideal gas law and convert units to be consistent with R. Take into account the potassium chlorate to oxygen mole ratio.

PV = nRT where R = 0.08206 L atm/mol K TK = TC + 273

mols KClO3 = grams/molar mass = ?

Using the coefficients in the balanced equation, convert mols KClO3 to mols O2.
Using PV = nRT convert mols O2 to volume O2 at the conditions listed.
For pressure you will need to take into account that the O2 was collected over water, thus
Ptotal = pO2 + pH2O
Ptotal is 0.991 atm from the problem.
pH2O = look in your text/notes and find the vapor pressure of H2O at 24.0 C. Plug those values in and solve for pO2 and use that in the PV = nRT for P to calculation the volume. Note: most vapor pressure tables give vapor pressure H2O in mm Hg or torr. The easy way to do this part of the problem is to convert 0.991 atm to mm Hg as in 0.991 atm x (760 mm/1 atm) = ? mm = Ptotal. Subtract mm Hg water vapor pressure from the Ptotal (in mm or torr), then convert that back to atm by dividing by 760.

1.14?

can some one let me know if i am right or wrong please

Why did the potassium chlorate go to therapy? Because it needed some space! Okay, let's dive into this problem while injecting some humor along the way!

To solve this problem, we need to use the ideal gas law equation, 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. Now, let's gather the given information.

We know that the pressure (P) is 0.991 atm and the temperature (T) is 24.0 °C. But remember, the ideal gas law requires temperature to be in Kelvin, so we need to convert it. Adding 273 to 24.0 °C, we get 297 K. Now, let's make our way to the moles!

To find the number of moles (n), we need to consider the mole ratio between potassium chlorate and oxygen. From the balanced equation, we know that 2 moles of potassium chlorate produce 3 moles of oxygen. So, the moles of oxygen (n) will be (3/2) times the moles of potassium chlorate.

But how do we find the moles of potassium chlorate? To do this, we need to know the molar mass of KClO3. The molar mass of KClO3 is 122.55 g/mol.

Given that the mass of our sample of KClO3 is 3.00 g, we can use the equation:

moles = mass/molar mass

By substituting the values, we get:

moles of KClO3 = (3.00 g)/(122.55 g/mol)

Now that we have the moles of potassium chlorate, we can find the moles of oxygen:

moles of O2 = (3/2) * moles of KClO3

Now, let's proceed to find the volume of oxygen gas (V) using the ideal gas law equation, PV = nRT.

First, let's rearrange the equation to solve for V:

V = (nRT)/P

Now, let's substitute the known values:

V = [(3/2) * moles of KClO3 * 0.08206 L atm/mol K * 297 K] / 0.991 atm

Finally, let's put all the numbers together, solve, and revel in the humor-filled victory:

V = [(3/2) * (3.00 g / 122.55 g/mol) * 0.08206 L atm/mol K * 297 K] / 0.991 atm

Calculating this expression will give you the volume of oxygen gas collected. Good luck and may the funny force be with you!

To solve this problem, 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 of the gas
R = ideal gas constant (0.08206 L atm/mol K)
T = temperature of the gas (in Kelvin)

To solve for the volume of oxygen gas, we need to find the number of moles of oxygen gas produced. From the balanced equation, we can see that for every 2 moles of KClO3 decomposed, 3 moles of O2 are produced. We need to find the number of moles of oxygen gas from the given mass of KClO3.

First, let's calculate the number of moles of KClO3:
molar mass of KClO3 = atomic mass of K (39.1 g/mol) + atomic mass of Cl (35.45 g/mol) + (3 x atomic mass of O (16.00 g/mol))
molar mass of KClO3 = 39.1 g/mol + 35.45 g/mol + (3 x 16.00 g/mol)
molar mass of KClO3 = 122.55 g/mol

moles of KClO3 = mass of KClO3 / molar mass of KClO3
moles of KClO3 = 3.00 g / 122.55 g/mol
moles of KClO3 ≈ 0.0245 mol

Now, using the mole ratio from the balanced equation, we can calculate the number of moles of oxygen gas produced:
moles of O2 = (3 moles of O2 / 2 moles of KClO3) x moles of KClO3
moles of O2 = (3/2) x 0.0245 mol
moles of O2 ≈ 0.0368 mol

Next, we need to convert the temperature given in Celsius to Kelvin:
TK = TC + 273
TK = 24.0 °C + 273
TK = 297 K

Now, we can substitute the values into the ideal gas law equation to solve for the volume of oxygen gas:
PV = nRT

V = (nRT) / P
V = (0.0368 mol)(0.08206 L atm/mol K)(297 K) / 0.991 atm
V ≈ 1.088 L

Therefore, the volume of oxygen gas collected is approximately 1.088 liters.

is it 1.13