Question

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!

Answer 1

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%