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#