Mass of Mg strip Temperature of water in the 600-mL beaker Vapor pressure of wat

ID: 1049666 • Letter: M

Question

Mass of Mg strip Temperature of water in the 600-mL beaker Vapor pressure of water at the temperature in the 600-mL beaker Volume of gas collected at room temperature Volume of gas in liters (V_2) Partial pressure of hydrogen gas at room temperature Partial pressure of hydrogen gas in atm (P_2) Volume of hydrogen gas at STP P_STP V_STP/T_STP = P_2V_2/T_2 (solve for V_STP) Moles of hydrogen produced based on amount of Mg used (use balanced equation) Molar volume of hydrogen (L/mol) at STP (calculated from your data) Average molar volume of hydrogen Molar volume of an ideal gas at STP Percent error

Explanation / Answer

I believe the equation you are studying is

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

Note that there is a 1:1 molar ratio between Mg reacted and H2 evolved.

Note that at STP, barometric pressure P = Pg + Pw where Pg is the partial pressure due to H2 and Pw is the partial pressure due to water vapour.

Trial 1

Trial 2

Trial 3

Mass of Mg strip (g)

0.080

0.079

0.078

Temperature of water in the 600 mL beaker (C)

21.0

21.5

22.0

Temperature in Kelvin scale (T2) (K)

294

294.5

295

Vapour pressure of water at the temperature in the 600 mL beaker (torr)

18.7

19.2

19.8

Volume of gas collected at room temperature (mL)

80.2

74.1

74.0

Volume of gas in liters (V2)

0.0802

0.0741

0.0740

Partial pressure of hydrogen gas at room temperature **1

741.3

740.8

740.2

Partial pressure of hydrogen gas in atmosphere (P2) **2

0.9754

0.9747

0.9739

Volume of hydrogen gas at STP

PSTPVSTP/TSTP = P2V2/T2 (L) **3

0.07929

0.07308

0.07280

Moles of hydrogen produced based on amount of Mg used (mole) **4

3.2922*10-3

3.2510*10-3

3.2099*10-3

Molar volume of hydrogen (L/mol) at STP **5

24.0842

22.4792

22.6798

Average molar volume of hydrogen (L/mol) **6

Molar volume of an ideal gas at STP (L/mol)

23.0811

22.414

Percent error

[(23.0811 – 22.414)L/mol]/(22.414 L/mol)*100 = 2.976 3.0

**1 P = 760 torr and Pw = 18.7; we know that P = Pg + Pw

===> 760 torr = Pg + 18.7 torr

===> Pg = (760 – 18.7) torr = 741.3 torr

**2 1 atm = 760 torr

====> 1 torr = (1 atm/760)

Therefore, 741.3 torr = (741.3 torr)*(1 atm/760 torr) = 0.9754 atm

**3 PSTP = 1 atm, TSTP = 298 K

VSTP = (P2V2/T2)*(TSTP/PSTP) = (0.9754 atm*0.0802 L/294 K)((298 K/1 am) = 0.07929 L

**4 Molar mass of Mg = 24.3 g/mol.

Mole(s) of Mg used = (0.080 g)*(1 mole/24.3 g) = 3.2922*10-3 mole.

Mole(s) of H2 produced = (3.2922*10-3 mole)*(1 mole H2/1 mole Mg) = 3.2922*10-3 mole.

**5 Molar volume = VSTP/mole of hydrogen produced = (0.07308 L)/(3.2510*10-3 mole) = 22.4792 mole.

**6 Average molar volume of hydrogen = (24.0842 + 22.4792 + 22.6798)L/mol/3 = 23.0811 L/mol