At a certain temperature, a 21.0-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 13.5 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients)?

SO2 +N2O<--->SO3+N2
Kc =  concentration   of  product /concentration  of  reactant
(SO3)(N2/(SO2)(N2O)
Find  the  concentration  of  each reactant  and  products
concentration =number  of  moles/  volume  in  liters
 moles  of SO3=  3.5/80=0.044moles
SO2=4.6/64=0.072 moles
N2=13.5/28=0.483  moles
N2O=0.98/44=0.022 moles
concentration  is  therefore=
SO3=0.044/21=  2.095  x  10^-3
SO2=0.072 /21=3.43 x  10^-3
N2=0.483/21=0.023
N2O=0.022/22= 1.048 x 10^-3

Kc   is  therefore  ={ (2.095 x10^-3)(0.023)}/ {(3.43 x10^-3)(1.048  x10^-3)}=13.40

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