Question 6ff77

Mar 26, 2016

Here's my take on this.

Explanation:

The idea here is that you are dealing with two distinct reactions, one in which carbon reacts with oxygen gas to produce carbon dioxide, ${\text{CO}}_{2}$, and one in which the two reactants produce carbon monoxide, $\text{CO}$.

You will have

${\text{C"_text((s]) + "O"_text(2(g]) -> "CO}}_{\textrm{2 \left(g\right]}}$

$\textcolor{red}{2} {\text{C"_text((s]) + "O"_text(2(g]) -> 2"CO}}_{\textrm{\left(g\right]}}$

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 $\text{750 mmHg}$ and temperature at ${18}^{\circ} \text{C}$. Use the ideal gas law equation to determine the molar volume of oxygen gas under these conditions

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{P V = n R T \implies \frac{V}{n} = \frac{R T}{P}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $\text{24.22 L}$.

Use elemental carbon's molar mass to determine how many moles of carbon you have in that $\text{11.2-g}$ 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 $\textcolor{red}{2} : 1$ mole ratio between carbon and oxygen gas, so when $y$ moles of carbon react to form carbon monoxide, they react with $\frac{1}{\textcolor{red}{2}} \cdot y$ 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 $\text{21.2 L}$ 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 $\textcolor{\mathmr{and} a n \ge}{\text{( * )}}$ to write

$x = 0.93248 - y$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\text{(* *)}}$ to find

$24.22 \cdot \left(0.93248 - y\right) + 12.11 y = 21.2$

$22.585 - 24.22 y + 12.11 y = 21.2$

$- 12.11 y = - 1.385 \implies y = \frac{1.385}{12.11} = 0.11437$

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 $\textcolor{red}{2} : 2$ mole ratio between carbon and carbon monoxide

${n}_{C {O}_{2}} = x = {\text{0.81811 moles CO}}_{2}$

${n}_{C O} = y = \text{0.11437 moles CO}$

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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\chi}_{C O} = \text{number of moles of CO"/"total number of moles}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you will have

${n}_{C O} = \left(0.11437 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))))/((0.11437 + 0.81811)color(red)(cancel(color(black)("moles}}}}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 0.123 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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, ${\text{Na"_2"CO}}_{3}$, and water

${\text{CO"_text(2(aq]) + color(purple)(2)"NaOH"_text((aq]) -> "Na"_2"CO"_text(3(aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

The sodium hydroxide solution is said to be $\text{2.5 N}$, or $\text{2.5 normal}$. 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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{normality" = "molarity" xx "no. of equivalents}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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 $\text{2.5-M}$ solution in a $\text{2-L}$ volume, which gives you

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${n}_{N a O H} = \text{2.5 M" * "2 L" = "5.0 moles NaOH}$

Assuming the all the moles of carbon dioxide will react, use the $1 : \textcolor{p u r p \le}{2}$ 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

${n}_{N a O H} = \text{5.0 moles" - "1.6362 moles" = "3.3638 moles NaOH}$

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

${c}_{N a O H} = \text{3.3638 moles"/"2 L" = "1.7 M}$

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, ${\text{H"_2"CO}}_{3}$, 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

${\text{CO"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "H"_2"CO}}_{\textrm{3 \left(a q\right]}}$

This tells you that one mole of carbon dioxide will actually provide two equivalents of protons, ${\text{H}}^{+}$, to the reaction, because carbonic acid is a diprotic acid.

If you start with $5.0$ equivalents of sodium hydroxide, the reaction will consume $2 \times 0.81811$ 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)|)))#