Question #91883

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Question #91883

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Answer 1 · 2015-11-04T11:14:14

$C_{4} H_{10}$ $"C"_4"H"_10$

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 $C_{x} H_{y}$ $"C"_x"H"_y$ ,

$C_{x} H_{y} + O_{2(g]} \to {CO}_{2(g]} + H_{2} O_{(l]}$ $"C"_x"H"_y + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])$

Balance this by using $x$ $x$ and $y$ $y$ . Since $x$ $x$ moles of carbon are present on the reactants' side, you need $x$ $x$ present on the products' side as well.

$C_{x} H_{y} + O_{2(g]} \to x {CO}_{2(g]} + H_{2} O_{(l]}$ $"C"_x"H"_y + "O"_text(2(g]) -> x"CO"_text(2(g]) + "H"_2"O"_text((l])$

You have $y$ $y$ moles of hydrogen on the reactants' side, and $2$ $2$ on the products' side, so multiply the water by $\frac{y}{2}$ $y/2$

$C_{x} H_{y} + O_{2(g]} \to x {CO}_{2(g]} + \frac{y}{2} H_{2} O_{(l]}$ $"C"_x"H"_y + "O"_text(2(g]) -> x"CO"_text(2(g]) + y/2"H"_2"O"_text((l])$

FInally, the number of moles of oxygen will be

$C_{x} H_{y} + (2 x + \frac{y}{2}) O_{2(g]} \to x {CO}_{2(g]} + \frac{y}{2} H_{2} O_{(l]}$ $"C"_x"H"_y + (2x + y/2)"O"_text(2(g]) -> x"CO"_text(2(g]) + y/2"H"_2"O"_text((l])$

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 ${1000 cm}^{3}$ $"1000 cm"^3$ of gas will represent a mixture of excess oxygen and carbon dioxide. Here is where the sodium hydroxide solution comes into play.

Sodium hydroxide solutions will absorb carbon dioxide to form potassium carbonate, $K_{2} {CO}_{3}$ $"K"_2"CO"_3$ , and water.

$2 {KOH}_{(aq]} + {CO}_{2(g]} \to K_{2} {CO}_{3(aq]} + H_{2} O_{(l]}$ $2"KOH"_text((aq]) + "CO"_text(2(g]) -> "K"_2"CO"_text(3(aq]) + "H"_2"O"_text((l])$

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 ${200 cm}^{3}$ $"200 cm"^3$ volume represents excess oxygen.

As a result, it follows that the reaction only needed

$V_{O_{2}} = {1500 cm}^{3} - {200 cm}^{3} = {1300 cm}^{3} O_{2}$ $V_(O_2) = "1500 cm"^3 - "200 cm"^3 = "1300 cm"^3" O"_2$

Moreover, you know that the reaction produced

$V_{C O_{2}} = {1000 cm}^{3} - {200 cm}^{3} = {800 cm}^{3} {CO}_{2}$ $V_(CO_2) = "1000 cm"^3 - "200 cm"^3 = "800 cm"^3" CO"_2$

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 $x=4$ and that you have

$(2x + y/2) = 13 ->$ 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, $"C"_4"H"_10$ .

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Question #91883

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