A piece of sodium metal undergoes complete reaction with water as follows: 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)?

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2017-10-18 1:15 pm

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A piece of sodium metal undergoes complete reaction with water as follows: 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g) The hydrogen gas generated is collected over water at 25.0oC. The volume of the gas is 246 mL measured at 1.00 atm. Calculate the number of grams of sodium used in the reaction. (Vapor pressure of water at 25.0oC = 23.8 torr.)

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戇戇居士

2017-10-18 2:52 pm

Consider the H₂ gas produced :
Pressure, P = [1 - (23.8/760)] atm
Volume, V = 246 mL = 0.246 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273.2 + 25.0) K = 298.2 K

Gas law: PV = nRT
Then, n = PV/(RT)

No. of moles of H₂ produced, n = [1 - (23.8/760)] × 0.246 / (0.08206 × 298.2) mol = 0.009738 mol

2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
Mole ratio Na : H₂ = 2 : 1
No. of moles of Na used = (0.009738 mol) × 2 = 0.01948 mol

Molar mass of Na = 23.0 g/mol
Mass of Na used = (0.1948 mol) × (23.0 g/mol) = 0.448 g

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