A student performed an experiment to calculate the molar volume of the hydrogen
ID: 568314 • Letter: A
Question
A student performed an experiment to calculate the molar volume of the hydrogen gas generated by the following reaction (see the experimental procedure given in this lab):
(Mg: 24.305 g/mol ; 1 Atm=101.325 kPa)
The student obtained the following results:
Volume of flask: 138.57 mL
Temperature of the water bath: 22 oC
Change in pressure (Pfinal-Pinitial): 10.8 kPa
Mass of Mg: 0.0149 g
The calculated volume of hydrogen produced (mL) and the molar volume of hydrogen at STP (L/mol), respectively will be:
13.9 mL; 22.9 L/molExplanation / Answer
Molar mass of Mg = 2.43 g/mol
Mg(s) + 2 HCl(aq) MgCl(aq) + H(g)
Mole ratio Mg : H = 1 : 1
No. of moles of Mg reacted = (0.0149 g) / (24.3 g/mol) = 0.000613 mol
No. of moles of H generated = 0.000613 mol
For the H generated :
Experiment conditions: P = 10.8 kPa, T = (273 + 22) K = 295 K, V = 138.57 mL = 0.13857 L
STP: P = 101.325 kPa, T = 273 K, V = ? L/mol
PV/T = PV/T
Hence, V = V × (P/P) × (T/T) = (0.13857 L) × (10.8/101.325) × (273/295) = 0.01367 L= 13.67 ml
Molar volume of H = (0.01367 L) / (0.000613 mol) = 22.3 L/mol
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