Balanced equation of the given reaction
2MoS2+7O2(g)→2MoO3(s)+4SO2(g)
Atomic masses considered
Mo→96 g/mol
S−32 g/mol
O→16 g/mol
Molar masses calculated
O2→2×16=32 g/mol
MoO3→96+3×16=144 g/mol
Mass of MoO3 to be produced
=1.82kg=1820g=1820g144gmol=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 MoO3 we require 7 moles O2(g)
So
To produce 12.64 moles of MoO3 we require
=72×12.64=18.96 molesO2(g)
Given
P→Pressure of O2(g)=715torr=715760atm
T→Temperature of O2(g)=38∘C=273+38=311K
n→No. of moles of O2(g)=18.96 moles
R→Universal gas constant=0.082LatmK−1mol−1
V→Volume of O2(g)=?
Inserting these values in the equation of state for ideal gas we get
PV=nRT
V=nRTP=18.96mol×0.082LatmK−1mol−1×311K715760atm
=513.95L
