Molar volume experiment with Mg ad HCl?

[匿名]

2017-10-09 8:39 pm

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)

Mg (s) + 2 HCl (aq) LaTeX: \longrightarrow⟶ MgCl2 (aq) + H2 (g)

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:

回答 (1)

戇戇居士

2017-10-09 10:33 pm

✔ 最佳答案

Molar mass of Mg = 2.43 g/mol

Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)
Mole ratio Mg : H₂ = 1 : 1
No. of moles of Mg reacted = (0.0149 g) / (24.3 g/mol) = 0.000613 mol
No. of moles of H₂ generated = 0.000613 mol

For the H₂ generated :
Experiment conditions: P₁ = 10.8 kPa, T₁ = (273 + 22) K = 295 K, V₁ = 138.57 mL = 0.13857 L
STP: P₂ = 101.325 kPa, T₂ = 273 K, V₂ = ? L/mol

P₁V₁/T₁ = P₂V₂/T₂
Hence, V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = (0.13857 L) × (10.8/101.325) × (273/295) = 0.01367 L

Molar volume of H₂ = (0.01367 L) / (0.000613 mol) = 22.3 L/mol

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