Answer to Question #237139 in Physical Chemistry for ruqwar

Question

sodium azide (Na3N) decomposes on heating to form sodium and nitrogen gas.

2Na3N--> 6Na+N2

1.35g of sodium nitride were heated.

a) calculate the mass of sodium formed on complete decomposition of the sodium nitride. give your answer to 3 significant figures.(AO2)

b)calculate the volume of nitrogen gas (in cm^3)produced on complete decomposition of 1.35g of sodium nitride at 320K and 110kpa. the gas constant R= 8.31JK^-1 mol^-1.

give your answer to 3 significant figures. (AO2)

Answer 1

The balanced reaction taking place is given as,

=> 2 Na₃N ------------> 6 Na + N₂ 

Given: Mass of Na₃N reacting = 1.35 g.

Molar mass of Na₃N = Atomic mass of Na X 3 + Atomic mass of N = 23 X 3 + 14 = 83 g/mol.

Since mass = moles X molar mass

=> 1.35 = moles of Na₃N reacting X 83

=> Moles of Na₃N reacting = 0.016265 mol approx.

a) From the above-balanced reaction we can see that 2 moles of Na₃N is reacting to produce 6 moles of Na.

Hence moles of N produced = moles of Na₃N reacting X 3 = 0.016265 X 3 = 0.0487952 mol approx.

Hence mass of Na produced = moles X atomic mass = 0.0487952 X 23 = 1.12 g.

Therefore the answer is 1.12 g.

b) Given: Temperature = 320 K

Pressure = 110 KPa = 1.0856 atm approx.                    (Since 1 atm = 101.325 KPa)

From the above balanced reaction we can see that 2 moles of Na₃N is reacting to produce 1 mole of N₂.

Hence moles of N₂ produced = moles of Na₃N reacting X 0.5 = 0.016265 X 0.5 = 0.0081325 mol approx.

According to ideal gas law,

=> PV = nRT

where P = pressure in atm = 1.0856 atm approx.

V = volume of gas in L

n = moles of gas = 0.0081325 mol approx.

R = gas constant = 0.0821 atm.L/K.mol

T = temperature in K = T in °C + 273 = 320 K

Hence substituting the values we get,

=> 1.0856 X V = 0.0081325 X 0.0821 X 320 

=> V = volume of gas produced = 0.197 L = 197 cm³ approx.           (Since 1 L = 1000 cm³ )

Therefore the answer is 197cm³ approx.