What volume of #O_2(g)# at 350. #"^o#C and a pressure of 5.25 atm is needed to completely convert 10.0 g of sulfur to sulfur trioxide?

Sulfur trioxide, #SO_3#, is produced in enormous quantities each year for use in the synthesis of sulfuric acid.
#S(s) + O_2(g) -> SO_2(g)#
#2SO_2 (g) + O_2 (g) -> 2SO_3(g)#

1 Answer
anor277
Nov 5, 2016

We can use the stoichiometric equation: #S(s) + 3/2O_2(g) rarrSO_3(g)#. Approx. #5*L# dioxygen gas are required.

Explanation:

#"Moles of sulfur"# #=# #(10.0*g)/(32.06*g*mol^-1)=0.312*mol#.

Given the stoichiometry, we require #3/2xx0.312*mol# of dioxygen gas, i.e. #0.468*mol#.

And now we solve for volume in the Ideal Gas equation:

#V=(nRT)/P# #=# #(0.468*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx623*cancel(K))/(5.25*cancel(atm))# #~=# #5*L#.

Note (i) that we converted #"degrees Celsius"# to #"degrees Kelvin"#, and (ii) that we had to use an appropriate #"Gas constant, R"#. #"R"# with various units would be supplied as supplementary material in any exam. The fact that our equation gave units of volume, when a volume was sought in the problem, can help to convince us that we got the problem right (for once!!).

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