Vapor Pressure of Water Aluminum carbide (Alreacts with water to produce methane

Question

Vapor Pressure of Water Aluminum carbide (Alreacts with water to produce methane gas according to the equation below. A sample containing Al is placed into a flask containing water. The gas produced is collected by the downward displacement of water. 380.0 ml of gas are collected at a temperature of 35.00°C and an atmospheric pressure of 730.8 torr. ( 43T (C) P (torrT (°C) P (torr) 4.6 9.2 12.8 17.5 23.8 31.8 42.2 0.0 40.0 50.0 10.0 15.0 20.0 25.0 30.0 35.0 55.3 92.5 60.0149.4 70.0233.7 80.0355.1 90.0525.8 100.0 760.0 A14C3 (s) + 12 H2O (l) 4 Al(OH)3 (s) + 3 CH4 (g) i. Calculate the pressure of CH4 gas in the collection space (in atm). i. How many grams of A43 were in the original sample ii. What is the mole fraction of CHgas in the final collected gas mixture? Atm grams Al4C3 4

Explanation / Answer

1.

Volume of gas = 380.0 mL = 0.380 L

Pressure of methane = 730.5 - 42.2 = 688.3 torr = 688.3 / 760 atm = 0.906 atm

2.

R = 0.0821 L.atm.K-1.mol-1

T = 35 + 273.15 = 308.15 K

Using ideal gas equation,

P V = n R T
n = 0.906 * 0.380 / (0.0821 * 308.15)

n = moles of methane = 0.0136 mol

From the balanced equation,

3 mol of methane is obtained from 1 mol of Al4C3

Then, 0.0136 mol of methane is obtained from 1 * 0.0136 / 3 = 0.00454 mol of Al4C3

Therefore,

Mass of Al4C3 = 17.6 g.

3.

According Raoult's law,

Partial vapour pressure = total pressure * mole fraction

688.3 = 730.5 * mole fraction of methane

Molefraction of methane = 688.3 / 730.5 = 0.942

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