#3.0 dm^3# of sulfur dioxide is reacted with #2.0 dm3# of oxygen according to the equation below. #(2SO_2)_(g) + (O_2)_(g) → (2SO_3)_(g)# What volume of sulfur trioxide (in #dm^3# ) is formed?

Assume the reaction goes to completion and all gases are measured at the same temperature and pressure.

1 Answer
Al E.
Jun 8, 2018

Consider,

#2SO_2(g) + O_2(g) to 2SO_3(g)#

Given,

#2.0"dm"^3 * (1.43"g")/"dm"^3 * "mol"/(32"g") approx 8.94*10^-2"mol"# of #O_2#, and

#3.0"dm"^3 * (2.62"g")/"dm"^3 * "mol"/(64.1"g") approx 0.123"mol"# of #SO_2#

Alternatively you could use the ideal gas law to derive the moles, but I used density because I could look them up (on an exam the former is probably your method of choice).

From the preceding data, we can conclude that #SO_2# is the limiting reactant. We need twice as much, according to the stoichiometry, and dividing it by two gives us #n_(SO_2) < n_(O_2)#. Hence,

#0.123"mol" * (2SO_3)/(2SO_2) * (80.1"g")/"mol" * "dm"^3/(2.62"g") approx 3.76"dm"^3# of #SO_3#

would ideally be produced.

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