10 #cm^3# of carbon monoxide and 10 #cm^3# of oxygen are mixed. What will be the total volume of gas present, in #cm^3#, after the reaction?
They are mixed and the volume is measured at the same temperature and pressure.
#2CO + O_2 -> 2CO_2#
a. 10
b. 15
c. 20
d. 25
The answer given is b. 15, why?
They are mixed and the volume is measured at the same temperature and pressure.
a. 10
b. 15
c. 20
d. 25
The answer given is b. 15, why?
1 Answer
The volume of gas will indeed be equal to
Explanation:
This one is a little tricky. And here's why I say that.
When it comes to reaction that involve gases kept under the same conditions for pressure and temperature, it's important to realize that the mole ratios that exist between the species involved in the reaction becomes equivalent to the volume ratio.
So, take a look at the balanced chemical equation for your reaction
#color(red)(2)"CO"_text((g]) + "O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g])#
Notice that you have a
But since we're working with gases under the same conditions for pressure an temperature, you can say the exact same thing about their volumes.
In other words, the reaction will always consume a volume of carbon monoxide that is twice as large as the volume of oxygen gas.
Now, here's where the tricky part comes in. Notice that the problem provides you with equal volumes of
This is a problem because you know that you'd need a volume of
You an thus conclude that you're dealing with a limiting reagent. More specifically, carbon monoxide will act as a limiting reagent because it will be consumed before all the oxygen gets a chance to react.
So, you can say that the reaction will only consume
#10color(red)(cancel(color(black)("cm"^3"CO"))) * ("1 cm"^3"O"_2)/(color(red)(2)color(red)(cancel(color(black)("cm"^3"CO")))) = "5 cm"^3color(white)(a)"O"_2#
So,
Now look at the
So, if
This means that after the reaction is finished, the reaction vessel will contain
#V_"gas" = overbrace(V_(O_2))^(color(purple)("in excess")) + overbrace(V_(CO_2))^(color(brown)("produced"))#
#V_"gas" = "5 cm"^3 + "1 0cm"^3 = color(green)("15 cm"^3)#