Question

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Molar Volume of a Gas Room temperature (T2): Barometric pressure: C be & torr Tria .0544 Data Trial 2 Trial 1 Mass of Mg strip Temperature of water in the 600-mL beaker 8.1 torr 142 Vapor pressure of water at the temperature in the 600-mL beaker torr torr S6.4 A.1 Volume of gas collected at room temperature mL mL ,67as 0541 S04L Volume of gas in liters (V2) L L Calculations 1. Partial pressure of hydrogen gas at room temperature torr torr torr Partial pressure of hydrogen gas in atm (P2) 2. atm atm atm 3. Volume of hydrogen gas at STP PST VSTP TSTP P>V;_(solve for VSTP) T2 = Moles of hydrogen produced based on amount of Mg used (use balanced equation) mol mol mol 4. L/mol Jmol Molar volume of hydrogen (L/mol) at STP (calculated from your data) L/mol 5. L/mol Average molar volume of hydrogen 6. L/mol Molar volume of an ideal gas at STP % 7. Percent error

Answer 1

Trial 1 :

Partial pressure of hydrogen gas at room temperature = barometric pressure - vapor pressure = 760.8 - 14.2 torr = 746.6 torr = Partial pressure of hydrogen gas at room temperature in atm , P2 = 746.6 torr = (1 torr = 0.001316 atm) = 0.982 atm

Pstp = 1 atm, Tstp = 273 K , Vstp = V

T1 = 22.3 + 273 = 295.3 K

V1 = 0.0795 lts

Pstp Vstp/Tstp = P2V2/T2

1 atm x V /273 = 0.982 atm x 0.0795 lts/295.3 K

V = 21.312/295.3 = 0.0722lts

Mg(s) + 2HCl(aq) ---> MgCl₂(aq) + H₂(g)

Moles of Magnesium = mass/molar mass = 0.078 g/24.31 g/mol = 0.0032 moles

Molar ratio of Mg and Hydrogen = 1:1

Moles of hydrogen = 0.0032 moles

Molar volume of hydrogen at STP = V/n = 0.0722 lts/0.0032 mol = 22.56 l/mol

Trial 2 :

Partial pressure of hydrogen gas at room temperature = barometric pressure - vapor pressure = 760.8 - 18.7 torr = 742.1 torr = Partial pressure of hydrogen gas at room temperature in atm , P2 = 742.1 torr = (1 torr = 0.001316 atm) = 0.976 atm

Pstp = 1 atm, Tstp = 273 K , Vstp = V

T1 = 22.3 + 273 = 295.3 K

V1 = 0.0541 lts

Pstp Vstp/Tstp = P2V2/T2

1 atm x V /273 = 0.976 atm x 0.0541 lts/295.3

V = 14.415/295.3 = 0.0488 lts

Mg(s) + 2HCl(aq) ---> MgCl₂(aq) + H₂(g)

Moles of Magnesium = mass/molar mass = 0.065 g/24.31 g/mol = 0.0027 moles

Molar ratio of Mg and Hydrogen = 1:1

Moles of hydrogen = 0.0027 moles

Molar volume of hydrogen at STP = V/n = 0.0488 lts/0.0027 mol = 18.07 l/mol

Trial 3 :

Partial pressure of hydrogen gas at room temperature = barometric pressure - vapor pressure = 760.8 - 18.7 torr = 742.1 torr = Partial pressure of hydrogen gas at room temperature in atm , P2 = 742.1 torr = (1 torr = 0.001316 atm) = 0.976 atm

Pstp = 1 atm, Tstp = 273 K , Vstp = V

T1 = 22.3 + 273 = 295.3 K

V1 = 0.0564 lts

Pstp Vstp/Tstp = P2V2/T2

1 atm x V /273 = 0.976 atm x 0.0564 lts/295.3

V = 15.028/295.3 = 0.0509 lts

Mg(s) + 2HCl(aq) ---> MgCl₂(aq) + H₂(g)

Moles of Magnesium = mass/molar mass = 0.0599 g/24.31 g/mol = 0.0025 moles

Molar ratio of Mg and Hydrogen = 1:1

Moles of hydrogen = 0.0025 moles

Molar volume of hydrogen at STP = V/n = 0.0509 lts/0.0025 mol = 20.36 l/mol

Average molar volume = 22.56 + 18.07 + 20.36)/3 = 60.99/3 = 20.33 l/mol

Theoretical value of molar volume of ideal gas at STP = 22.4 l/mol

Percent error = ([exp value - theoretical value])/theoretical value) x 100 = ([20.33-22.4)/22.4) x 100 = 9.24 %