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

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  1. 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.

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  2. is it 1.13

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  3. 1.14?

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

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