Answer to Question #111344 in General Chemistry for Alana M

Question

At what pressure (in atm) is the nitrogen gas sample that is collected when 48.4 g of NaN3 decomposes? The temperature of the gas is 25oC and the volume is 18.4 L. Round your answer to three sig figs.

2NaN3 (s) --> 2 Na (s) + 3 N2 (g)

Answer 1

Firstly, we need to calculate how many moles of N₂ is formed in chemical reaction

(2NaN₃ (s) --> 2 Na (s) + 3 N₂ (g))

In order to do this we need to calculate how many moles of NaN₃ has reacted:

M(NaN₃) = 22.99 g/mol + 3*14.007 g/mol = 65.011 g/mol - Molar mass calculated from Periodic tanle

n(NaN₃) = m/M = 48.4 g/65.011 g/mol = 0.745 mol (of NaN3 has reacted)

According to reaction amount of moles of N₂ is:

n(N₂) = (3*0.745 mol)/2 = 1.118 mol - amount of moles of N₂

Now using Mendeleev-Clapeyron equation (PV=nRT, P - pressure, V - volume, n - amount of substance (mol), R- gas constant (8.314 J/mol*K) T- temperature) it's possible to calculate pressure of N₂ gas:

P=nRT/V

P = (1.118 mol * 8.314 J/mol*K * 298 K)/48.4 L = 150.539429 J/L = 150539.429 Pa (It's known that

1J/L = 1000 Pa) - pressure of N₂ gas (Pa)

25^oC was transformed to Kelvin scale (25+273 = 298 K)

Finally, it's known that 1 atm is 101325 Pa. Therefore 150539.429 Pa is 150539.429/101325=1.486 atm.

Answer: pressure of N₂ gas is 1.486 atm.