Under conditions of #"NTP"#, what volume of #SO_2(g)# results from burning a #10*g# mass of elemental sulfur?

1 Answer
anor277
May 19, 2017

We address the stoichiometric equation...........and I get a volume of under #10*L#...............

Explanation:

#S(s)+O_2(g) rarrSO_2(g)#

Now depending on your definition of #"NTP"# (they seem to change with the wind across syllabuses) we can work out the volume. I use values of #"NTP"# to specify #P=1*atm#, and #T=293.15*K#.

#"Moles of sulfur "-=" Moles of sulfur dioxide"#

#"Moles of sulfur"-=(10*g)/(32.06*g*mol^-1)=0.312*mol#.

And thus there are #0.312*mol# #SO_2# gas........given conversion stoichiometric conversion.

#V=(nRT)/P=(0.312*molxx0.0821*(L*atm)/(K*mol)xx293.15*K)/(1*atm)#

#=7.5*L#

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