periental Data Trial 1 Trial 2 (a) Mass of Magnesium metal used (b) Volume of H2

ID: 699200 • Letter: P

Question

periental Data Trial 1 Trial 2 (a) Mass of Magnesium metal used (b) Volume of H2 gas collected (c) Temperature of H2 gas collected (d) Atmospheric Pressure (e) Temperature of Water in bath (bucket) (f) Vapor Pressure of Water at above temperature O.03 0.O34 3,12 mL 38.50ML 75%, 3 mmHg | 75X-3 mm 25.0° C 22 21-1 224 mmHg wwnH Data Analysis Using your experimental data, determine the value of R, the gas constant. Show all your conversions and calculations for each step clearly in the table below. Pay attention to units and significant figures.

Explanation / Answer

The balanced chemical equation for the reaction is given as

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

As per the stoichiometry of the reaction,

1 mole Mg = 1 mole H2

Trial 1

Trail 2

(a) Mass of magnesium metal used (g)

0.039

0.034

(b) Mole(s) of magnesium metal used = (mass of magnesium metal)/(atomic mass of magnesium) (mole)

[Atomic mass of magnesium = 24.305 g/mol]

(0.039)*(1/24.305) = 0.001605

(0.034)*(1/24.305) = 0.001399

0.001605

0.001399

(d) Volume of H2 gas collected (mL)

43.12

38.50

(e) Volume of H2 gas collected in L = (volume in mL)*(1 L/1000 mL)

(43.12)*(1/1000) = 0.04312

(38.50)*(1/1000) = 0.03850

(f) Temperature of H2 gas collected (°C)

23.6

23.9

(g) Temperature of H2 gas collected in K = (temperature in °C + 273)

(23.6 + 273) = 296.6

(23.9 + 273) = 296.9

(h) Atmospheric Pressure (mmHg)

758.3

(i) Temperature of water in bath (bucket) (°C)

25.0

21.1

(j) Vapor Pressure of Water at above temperature

22.4

(k) Pressure of dry H2 gas = (atmospheric pressure) – (vapor pressure of water) (mmHg)

(758.3 – 22.4) = 735.9

(l) Pressure of dry H2 gas in atm = (pressure in mmHg)*(1 atm/760 mmHg)

(735.9)*(1/760) = 0.9683

(m) Gas constant, R = (pressure in atm)*(volume in L)/(moles of H2).(temperature in K) (L-atm/mol.K)

(0.9683)*(0.04312)/(0.001605).(296.6) = 0.0877

(0.9683)*(0.03850)/(0.001399).(296.9) = 0.0897

(n) Mean value of R (L-atm/mol.K)

½*(0.0877 + 0.0897) = 0.0887