Question

Question #6ff77

Answer 1

Here's my take on this.

Explanation:

!! EXTREMELY LONG ANSWER !!

The idea here is that you are dealing with two distinct reactions, one in which carbon reacts with oxygen gas to produce carbon dioxide, `"CO"_2`, and one in which the two reactants produce carbon monoxide, `"CO"`.

You will have

`"C"_text((s]) + "O"_text(2(g]) -> "CO"_text(2(g])`

`color(red)(2)"C"_text((s]) + "O"_text(2(g]) -> 2"CO"_text((g])`

Now, the trick here is to use the concept of molar volume of a gas, i.e. the volume occupied by one mole of an ideal gas under specific conditions for pressure and temperature.

In your case, pressure is set at `"750 mmHg"` and temperature at `18^@"C"`. Use the ideal gas law equation to determine the molar volume of oxygen gas under these conditions

`color(purple)(|bar(ul(color(white)(a/a)color(black)(PV = nRT implies V/n = (RT)/P)color(white)(a/a)|)))`

Do not forget to convert the pressure from mmHg to atm and the temperature from degrees Celsius to Kelvin

`V/n = (0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 18)color(red)(cancel(color(black)("K"))))/(750/760color(red)(cancel(color(black)("atm")))) = "24.22 L mol"^(-1)`

So, under these conditions for pressure and temperature, one mole of oxygen gas occupies exactly `"24.22 L"`.

Use elemental carbon's molar mass to determine how many moles of carbon you have in that `"11.2-g"` sample

`11.2 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "0.93248 moles C"`

Let's assume the `x` represents the number of moles of carbon that react to produce carbon dioxide and `y` represents the number of moles of carbon that reacts to produce carbon monoxide.

You know that

`x + y = "0.93248 moles" " " " "color(orange)("( * )")`

Now focus on the oxygen. Notice that you have a `1:1` mole ratio between carbon and oxygen gas in the first equation.

This means that when `x` moles of carbon react to form carbon dioxide, they react with `x` moles of oxygen gas.

Use the molar volume of the gas to write the number of moles of oxygen gas in terms of the volume it would occupy

`x color(red)(cancel(color(black)("moles O"_2))) * "24.22 L"/(1color(red)(cancel(color(black)("mole O"_2)))) = (24.22 * x)color(white)(a)"L"`

Do the same for the second reaction, This time, you have a `color(red)(2):1` mole ratio between carbon and oxygen gas, so when `y` moles of carbon react to form carbon monoxide, they react with `1/color(red)(2) * y` moles of oxygen.

This means that you have

`1/color(red)(2)y color(red)(cancel(color(black)("moles O"_2))) * "24.22 L"/(1color(red)(cancel(color(black)("mole O"_2)))) = (12.11 * y)color(white)(a)"L"`

However, you know that you have `"21.2 L"` of oxygen gas available. This means that you can write

`(24.22 * x)"L" + (12.11 * y)"L" = "21.2 L"" " " "color(orange)("(* *)")`

Use equation `color(orange)("( * )")` to write

`x = 0.93248 - y`

Plug this into equation `color(orange)("(* *)")` to find

`24.22 * (0.93248 - y) + 12.11y = 21.2`

`22.585 - 24.22y + 12.11y = 21.2`

`-12.11y = -1.385 implies y = 1.385/12.11 = 0.11437`

You will thus have

`x = 0.93248 - 0.11437 = 0.81811`

So, you know that the reaction will produce - keep in mind that you have a `color(red)(2):2` mole ratio between carbon and carbon monoxide

`n_(CO_2) = x = "0.81811 moles CO"_2`

`n_(CO) = y = "0.11437 moles CO"`

The mole fraction of carbon monoxide in the resulting mixture will be

`color(purple)(|bar(ul(color(white)(a/a)color(black)(chi_(CO) = "number of moles of CO"/"total number of moles")color(white)(a/a)|)))`

In your case, you will have

`n_(CO) = (0.11437 color(red)(cancel(color(black)("moles"))))/((0.11437 + 0.81811)color(red)(cancel(color(black)("moles")))) = color(green)(|bar(ul(color(white)(a/a)0.123color(white)(a/a)|)))`

I'll leave the answer rounded to three sig figs.

Now, carbon dioxide / carbon monoxide this mixture is passed through a sodium hydroxide solution. The carbon dioxide will react with the sodium hydroxide to form aqueous sodium carbonate, `"Na"_2"CO"_3`, and water

`"CO"_text(2(aq]) + color(purple)(2)"NaOH"_text((aq]) -> "Na"_2"CO"_text(3(aq]) + "H"_2"O"_text((l])`

The sodium hydroxide solution is said to be `"2.5 N"`, or `"2.5 normal"`. Normality is defined as number of equivalents of solute per liter of solution.

A solution's normality depends on the reaction that takes place. To keep things simple, you can use the molarity of the solution instead. Keep in mind that you have

`color(purple)(|bar(ul(color(white)(a/a)color(black)("normality" = "molarity" xx "no. of equivalents")color(white)(a/a)|)))`

Here one mole of sodium hydroxide provides one equivalent of hydroxide anions to the reaction, so molarity and normality are interchangeable.

This means that you have a `"2.5-M"` solution in a `"2-L"` volume, which gives you

`color(blue)(|bar(ul(color(white)(a/a)c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution"color(white)(a/a)|)))`

`n_(NaOH) = "2.5 M" * "2 L" = "5.0 moles NaOH"`

Assuming the all the moles of carbon dioxide will react, use the `1:color(purple)(2)` mole ratio that exists between aqueous carbon dioxide and sodium hydroxide to determine how many moles of the latter are consumed by the reaction

`0.81811 color(red)(cancel(color(black)("moles CO"_2))) * (color(purple)(2)color(white)(a)"moles NaOH")/(1color(red)(cancel(color(black)("mole CO"_2)))) = "1.6362 moles NaOH"`

You will be left with

`n_(NaOH) = "5.0 moles" - "1.6362 moles" = "3.3638 moles NaOH"`

Assuming that the volume remains unchanged, the molarity of the sodium hydroxide solution will be

`c_(NaOH) = "3.3638 moles"/"2 L" = "1.7 M"`

Since this is equivalent to the solution's normality, you will have

`"N" = color(green)(|bar(ul(color(white)(a/a)"1.7 normal"color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs.

SIDE NOTE

It's worth noting that you're actually dealing with a neutralization reaction here. Aqueous carbon dioxide exists in equilibrium with carbonic acid, `"H"_2"CO"_3`, so a better description of what is going here would be

`overbrace("H"_2"CO"_text(3(aq]))^(color(red)("CO"_text(2(aq]) + "H"_2"O"_text((l]))) + color(purple)(2)"NaOH"_text((aq]) -> "Na"_2"CO"_text(3(aq]) + "H"_2"O"_text((l])`

Notice that you have

`"CO"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "H"_2"CO"_text(3(aq])`

This tells you that one mole of carbon dioxide will actually provide two equivalents of protons, `"H"^(+)`, to the reaction, because carbonic acid is a diprotic acid.

If you start with `5.0` equivalents of sodium hydroxide, the reaction will consume `2 xx 0.81811` equivalents of sodium hydroxide, since that's how many equivalents of carbon dioxide you have.

Once again, the answer would be

`"N" = ((5.0 - 2 xx 1.6362)color(white)(a)"equiv.")/"2 L" = color(green)(|bar(ul(color(white)(a/a)"1.7 normal"color(white)(a/a)|)))`