Question

A sample of NaN3(s) fully decomposed into Na(s) and N2(g), and the resulting gas was collected over water at 52°C. The collected gas occupied 2.0 L at a pressure of 690 torr. If the vapor pressure of water is 100 torr at 52°C, what mass of NaN3(65.02 g/mol) was reacted?

(1) 2.5 g (2) 2.9 g (3) 3.4 g (4) 3.8 g (5) 4.4 g

Answer 1

Ans. option 1 i.e.,2.51g

applying ideal gas equation PV=nRT.

P=pressure of N₂ gas but total pressure given in question is 690torr i.e., moist gas along with watervapour.

so, to get pressure shown by only N₂ gas we subract watervapour pressure from total pressure

P shown by N₂gas = 690-100=590 torr

1atm -----> 760torr

_atm------> 590 torr apply cross multiplication we get 590x1/760 = 0.776atm

V= 2lit occupied by N₂ gas. n= Wt of N₂/Mwt of N₂ =Wt of N₂/28, R= 0.0821lit.atm/mol k, T=52+273 =325

apply PV=nRT 0.776x2=(Wt of N₂/28) x0.0821x325

Wt of N₂ = 0.776x2x28/0.0821x325 =1.628gm (1.628gm of N₂ gas collected over water given by NaN₃)

2NaN₃--------> 2Na + 3N₂

   130gm NaN₃-------> 3x28gm N₂

___gm NaN₃-------> 1.628gmN₂   applying cross multiplication

130x1.628/2x28 = 2.51 gm of NaN₃