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
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.