Balanced equation of the given reaction
#2 MoS_2 + 7 O_2(g) ->2 MoO_3 (s) + 4 SO_2(g)#
Atomic masses considered
#Mo->96 " g/mol"#
#S-32 " g/mol"#
#O->16 " g/mol"#
Molar masses calculated
#O_2->2xx16=32 " g/mol"#
#MoO_3->96+3xx16=144 " g/mol"#
Mass of #MoO_3# to be produced
#=1.82kg=1820g=(1820g)/(144g/"mol")=12.64mol#
By the stochiometric ratio of reactants and products in the balanced equation of the reaction involved we can say
To produce 2 moles of #MoO_3# we require 7 moles #O_2(g)#
So
To produce 12.64 moles of #MoO_3# we require
# =7/2xx12.64=18.96" moles" O_2(g)#
Given
#P-> "Pressure of "O_2(g)=715"torr"=715/760atm#
#T-> "Temperature of "O_2(g)=38^@C=273+38=311K#
#n-> "No. of moles of "O_2(g)=18.96 " moles"#
#R->"Universal gas constant"=0.082LatmK^-1mol^-1#
#V-> "Volume of "O_2(g)=?#
Inserting these values in the equation of state for ideal gas we get
#PV=nRT#
#V=(nRT)/P=(18.96molxx0.082LatmK^-1mol^-1xx311K)/(715/760atm)#
#=513.95L#
