Given:
Room temperature: 293.0 K
Barometric pressure: 764.0 mmHg
Vapor of water: 17.5 mmHg
Volume of O2 collected: 68.00 mL
Density of H2O2: 1.01 g/mL
% Composition H2O2: 3.02 %
Volume of H2O2 used: 5.00 mL
Letter of the unknown solution of H2O2: A
Volume of O2 collected for the unknown: 43.00 mL
Calculate the corrected barometric pressure. (mmHg)
Calculate the volume of O2 at STP. (mL)
Based on the reaction stoichiometry, calculate the number of moles of O2. (moles)
Calculate the mass of O2: (g)
Calculate molar volume at STP: (L/mole)
Calculate the % error of the molar volume at STP.
Calculate R in the ideal gas law. (atm L mole-1 K-1)
Calculate the % error of the value of R.
Calculate the % composition of H2O2 in the unknown solution. (Round your answer to the nearest integer)
Someone please help!
the corrected barometric pressure. (mmHg) = Barometric pressure - Vapor of water pressure = 764 - 17.5 = 746.5 mmHg = 0.982 atm (760 mmHg = 1 atm)
volume of O2 at STP , VSTP = V Tstp Po2 /T Pstp = 0.043 x 273.15 x 746.5/293 x 760 = 8767.978425/222680 = 0.0394 lts = 39.4 ml
2H2O2 ---> 2H2O + O2
Molar ratio of H2O2 and O2 = 2:1
Ideal gas equation : PV = nRT
0.982 atm x 0.043 lts = n x 0.0821 lt atm/mol K x 293 K
0.042226/24.0553 = n
n = 0.00176 moles
Number of moles of hydrogen peroxide = 2 x 0.00176 = 0.00352 moles
mass of oxygen = moles x molar mass = 0.00176 x 32 g/mol = 0.056 g
molar volume at STP: (L/mole) = 0.0394 lts/0.00176 moles = 22.39 l/mol
Theoretical molar volume = 22.4 l/mol
Percent error of the molar volume at STP = [Theoretical value - practical value]/theoretical value ) x 100 = 0.01/22.4) x 100 = 0.045 %
PV = nRT
0.982 atm x 0.043 lts = 0.00176 x R x 293 K
0.042226/0.51568 = R
R = 0.0819 lt atm/mol K
Percent error = [0.0821 -0.0819]/0.0821) x 100 = 0.0002/0.0821) x 100 = 0.24%
Number of moles of hydrogen peroxide = 0.00352 moles
Mass of hydrogen peroxide = 0.00352 mol x 122.51 g/mol = 0.431 g
Density = 1.01 g/ml
Volume of hydrogen peroxide = 5 ml
Initial mass of hydrogen peroxide used = density x volume = 5 ml x 1.01 g/ml = 5.05 g
Percent composition of H2O2 in the unknown solution : (0.431 g/5.05 g) x 100 = 8.5 % = 9%
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