J.R. S. answered • 02/28/21

Ph.D. University Professor with 10+ years Tutoring Experience

2NaN_{3} ==> 2Na + 3N_{2}

Total pressure = 733.8 mm Hg

Vapor pressure of H_{2}O @ 26ºC = 25.2 mm Hg

Vapor pressure due to N_{2} gas collected over water = 733.8 mm Hg - 25.2 mm Hg = 708.6 mm Hg

Converting this to atm we have 708.6 mm Hg x 1 atm/760 mm Hg = __0.932 atm = P__

Converting this to liters of N_{2}, we use the Ideal gas law: PV = nRT

To find moles N_{2}, we use the stoichiometry of the balanced equation:

29.4 g NaN_{3} x 1 mol / 65 g x 3 mol N_{2} / 2 mol NaN_{3} = __0.302 moles N___{2}__ = n__

R = gas constant = 0.0821 Latm/Kmol

T = temp in K = 26 + 273 = 299K

Solving for V, we have...

V = nRT/P = (0.302)(0.0821)(299) / 0.932

V = 7.95 L

J.R. S.

02/28/21

Dawson H.

It says that answer is wrong, so I don't get it at all.02/28/21