Question #8c2bd
1 Answer
Here's what I got.
Explanation:
The balanced chemical equation that describes this reaction looks like this
#2"CO"_ ((g)) + "O"_ (2(g)) -> 2"CO"_ (2(g))#
Notice that the reaction consumes
Now, you know that all three gases are kept under the same conditions for pressure and temperature. This means that you can treat the mole ratios that exist between the three gases as volume ratios.
In other words, you can say that the reaction consumes
So, in order for the reaction to consume all
#50 color(red)(cancel(color(black)("cm"^3color(white)(.)"O"_2))) * ("2 cm"^3color(white)(.)"CO")/(1color(red)(cancel(color(black)("cm"^3color(white)(.)"O"_2)))) = "100 cm"^3color(white)(.)"CO"#
Since you don't have enough carbon monoxide to ensure that all the sample of oxygen gas can take part in the reaction
#overbrace("100 cm"^3color(white)(.)"CO")^(color(blue)("what you need")) " " > " " overbrace("80 cm"^3color(white)(.)"CO")^(color(blue)("what you have"))#
you can say that carbon monoxide will act as a limiting reagent, i.e. it will be completely consumed before all the sample of oxygen gas will get the chance to react.
Therefore, you can say that the reaction will consume
#80 color(red)(cancel(color(black)("cm"^3color(white)(.)"CO"))) * ("1 cm"^3color(white)(.)"O"_2)/(2color(red)(cancel(color(black)("cm"^3color(white)(.)"CO")))) = "40 cm"^3color(white)(.)"O"_2#
and produce
#80 color(red)(cancel(color(black)("cm"^3color(white)(.)"CO"))) * ("2 cm"^3color(white)(.)"CO"_2)/(2color(red)(cancel(color(black)("cm"^3color(white)(.)"CO")))) = "80 cm"^3color(white)(.)"CO"_2#
Once the reaction is complete, you will be left with
#"50 cm"^3color(white)(.)"O"_2 - "40 cm"^3color(white)(.)"O"_2 = "10 cm"^3color(white)(.)"O"_2#
The reaction vessel will contain
#{( "80 cm"^3color(white)(.)"CO"_2 color(white)(a)-> "produced by the reaction"), ("10 cm"^3color(white)(.)"O"_2 color(white)(aa)-> "unreacted") :}#