Question

(A.) Phosgene is a deadly gas made up of carbon, oxygen and chlorine. Its density at 1.05 atm and 25C. is 4.24 g/L. Show that the molecular formula of phossgene is COCl₂ if it's compostion is 12.1 % carbon, 16.2% oxygen, and 71.7% chlorine by mass.

(B.) When exposed to UV light, phosgene, decomposes according to the reaction

COCl₂(g)------>CO(g)+Cl₂(g)

All the air removed from a 44.8 L flask. The flask is then filled the phosgene to a pressure of 1.0 atm at temperatue 0.C. The flask is selaed and exposed to UV light. What are the partial pressure of the CO and Cl₂ after the decompostion of phosgene is complete?

Pls show step by step on how each one is solved thank you in advance :)

Answer 1

A) C = 12.1/12 = 1.008

    Cl = 71.7/35.5 = 2.02

    O = 16.2 /16 = 1.0125

simplest ratio

C = 1.008/1.008 = 1

cl = 2.02/1.008 = 2

O = 1.0125/1.008 = 1

empirical formula = COCl2

empirical formula mass = 12+16+2*35.5 = 99 g/mol

mOlarmass of phosgene = DRT/

                       = 4.24*0.0821*298/1.05

                       = 98.8 g/mol

molecular formula = COCl2

B) No of mol of COCl2 filled (n) = PV/RT

            = 1*44.8/(0.0821*273.15)

            = 2 mol

from equation, 1 mol COCl2(g) = 1mol CO(g) = 1 mol Cl2(g)

partial pressure of CO = 2*0.0821*273.15/44.8 = 1 atm

partial pressure of Cl2 = 2*0.0821*273.15/44.8 = 1 atm