Question #91883
1 Answer
Explanation:
The important thing to realize here is that all volumes were measured under the same conditions for pressure and temperature.
This means that the volume ratios that exists between the gaseous compounds is equivalent to the mole ratios that exist between the same species.
It's important to set up a generic balanced chemical equation for the combustion of your hydrocarbon, which we'll name
#"C"_x"H"_y + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])#
Balance this by using
#"C"_x"H"_y + "O"_text(2(g]) -> x"CO"_text(2(g]) + "H"_2"O"_text((l])#
You have
#"C"_x"H"_y + "O"_text(2(g]) -> x"CO"_text(2(g]) + y/2"H"_2"O"_text((l])#
FInally, the number of moles of oxygen will be
#"C"_x"H"_y + (2x + y/2)"O"_text(2(g]) -> x"CO"_text(2(g]) + y/2"H"_2"O"_text((l])#
Now, another important thing to realize here is that oxygen is present in excess. Notice that the reaction produces carbon dioxide, a gas, and liquid water.
This means that the remaining
Sodium hydroxide solutions will absorb carbon dioxide to form potassium carbonate,
#2"KOH"_text((aq]) + "CO"_text(2(g]) -> "K"_2"CO"_text(3(aq]) + "H"_2"O"_text((l])#
The actual reaction is not really important here, because you're supposed to assume that all the carbon dioxide produced by the reaction was absorbed by the potassium hydroxide solution.
This means that the final
As a result, it follows that the reaction only needed
#V_(O_2) = "1500 cm"^3 - "200 cm"^3 = "1300 cm"^3" O"_2#
Moreover, you know that the reaction produced
#V_(CO_2) = "1000 cm"^3 - "200 cm"^3 = "800 cm"^3" CO"_2#
which was completely absorbed by the potassium hydroxide solution.
Now, the volume ratios are equivalent to the mole ratio. This means that
#n_"hydrocarbon"/n_(CO_2) = V_"hydrocarbon"/V_(CO_2) = (200color(red)(cancel(color(black)("cm"^3))))/(800color(red)(cancel(color(black)("cm"^3)))) = 1/4#
and
#n_"hydrocarbon"/n_(O_2) = V_"hydrocarbon"/V_(O_2) = (200color(red)(cancel(color(black)("cm"^3))))/(1300color(red)(cancel(color(black)("cm"^3)))) = 2/13#
This means that the balanced chemical equation becomes
#"C"_x"H"_y + overbrace((2x + y/2))^(color(blue)(=13/2))"O"_text(2(g]) -> overbrace(x)^(color(blue)(=4))"CO"_text(2(g]) + y/2"H"_2"O"_text((l])#
#"C"_x"H"_y + 13/2"O"_text(2(g]) -> 4"CO"_text(2(g]) + y/2"H"_2"O"_text((l])#
Notice that
#(2x + y/2) = 13 -># the number of atoms of oxygen
This means that you get
#2 * 4 + y/2 = 13#
#16 + y = 26 implies y = 10#
Therefore, the balanced chemical equation will be
#"C"_4"H"_10 + 13/2"O"_text(2(g]) -> 4"CO"_text(2(g]) + 5"H"_2"O"_text((l])#
The hydrocarbon is butane,