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Sodium metal reacts with water to produce hydrogen gas according to the following equation:rnrn2na(s) + 2h2o(l) → 2naoh(aq) + H2(g)rnrnThe product gas, H2, is collected over water at a temperature of 25 °c and a pressure of 752.0 mm Hg. if the wet h2 gas formed occupies a volume of 7.43 l, the number of moles of na reacted was mol. The vapor pressure of water is 23.8 mm hg at 25 °c.
Accepted Answer
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
At 25 °C, the vapor pressure of water is 23.8 mm Hg.
Therefore, the partial pressure of H2 in the collected gas is:
P(H2) = Total pressure - Vapor pressure of water
P(H2) = 752.0 mm Hg - 23.8 mm Hg
P(H2) = 728.2 mm Hg
Using the Ideal Gas Law and the partial pressure of H2, we can calculate the number of moles of H2 produced:
PV = nRT
n = PV/RT
n = (728.2 mm Hg * 7.43 L) / (0.0821 L atm / mol K * 298 K)
n = 3.00 mol
Since 1 mol of H2 is produced for every 2 mol of Na reacted, the number of moles of Na reacted is:
n(Na) = n(H2) / 2
n(Na) = 3.00 mol / 2
n(Na) = 1.50 mol
Therefore, the number of moles of Na reacted is 1.50 mol.
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