Question #8f087

1 Answer
Dec 26, 2017


#"0.5 dm"^3#


The idea here is that if all three gases are kept under the same conditions for pressure and temperature, then you can treat the mole ratios that exist between them as volume ratios.

#2"SO"_ (2(g)) + "O"_ (2(g)) -> 2"SO"_ (3(g))#

So if the conditions for pressure and temperature are the same for all three gases, you can say that in order for the reaction to produce #"3 dm"^3# of sulfur trioxide it must consume #"2 dm"^3# of sulfur dioxide and #"1 dm"^3# of oxygen gas.

Now, your goal here is to figure out if you're dealing with a limiting reagent. To do that, pick the volume of a reactant and check to see if you have enough of the second reactant to ensure that the first reactant is completely consumed by the reaction.

Let's pick sulfur dioxide this time. You have

#0.5 color(red)(cancel(color(black)("dm"^3color(white)(.)"SO"_2))) * ("1 dm"^3color(white)(.)"O"_2)/(2color(red)(cancel(color(black)("dm"^3color(white)(.)"SO"_2)))) = "0.25 dm"^3color(white)(.)"O"_2#

This means that in order for the reaction to consume #"0.5 dm"^3# of sulfur dioxide, ti must also consume #"0.25 dm"^3# of oxygen gas.

Since you have more oxygen gas available

#overbrace("1 dm"^3color(white)(.)"O"_2)^(color(blue)("what you have")) " " > " " overbrace("0.25 dm"^3color(white)(.)"O"_2)^(color(blue)("what you need"))#

you can say that oxygen gas is in excess. This is equivalent to saying that sulfur dioxide is the limiting reagent, i.e. it will be completely consumed before all the volume of oxygen gas will be able to take part in the reaction.

So, you know that the reaction consumes #"0.5 dm"^3# of sulfur dioxide and #"0.25 dm"^3# of oxygen gas, so you can say that it will produce

#0.5 color(red)(cancel(color(black)("dm"^3color(white)(.)"SO"_2))) * ("2 dm"^3 color(white)(.)"SO"_3)/(2color(red)(cancel(color(black)("dm"^3color(white)(.)"SO"_2)))) = color(darkgreen)(ul(color(black)("0.5 dm"^3color(white)(.)"SO"_3)))#

The answer is rounded to one significant figure.