Question a20b7

Jun 18, 2017

It would require 80 g of $\text{NaOH}$.

Explanation:

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 40.00$
$\textcolor{w h i t e}{m m l l} \text{2NaOH" + "CO"_2 → "Na"_2"CO"_3 + "H"_2"O}$

Calculate the moles of $\text{NaOH}$

$\text{Moles of NaOH" = 20 color(red)(cancel(color(black)("g NaOH"))) × "1 mol NaOH"/(40.00 color(red)(cancel(color(black)("g NaOH")))) = "0.500 mol NaOH}$

Calculate the moles of ${\text{CO}}_{2}$ in the mixture

${\text{ Moles of CO"_2 = 0.500 color(red)(cancel(color(black)("mol NaOH"))) × "1 mol CO"_2/(2 color(red)(cancel(color(black)("mol NaOH")))) = "0.250 mol CO}}_{2}$

Calculate the moles of $\text{CO}$ in the mixture

$\text{Moles of CO + Moles of CO"_2 = "total moles}$

$\text{Moles of CO + 0.250 mol = 1 mol}$

$\text{Moles of CO = (1 - 0.250) mol = 0.750 mol}$

Write the balanced equation for the oxidation of $\text{CO}$

${\text{2CO + O"_2 → "2CO}}_{2}$

${\text{Moles of CO"_2color(white)(l) "from CO" = 0.750 color(red)(cancel(color(black)("mol CO"))) × "2 mol CO"_2/(2 color(red)(cancel(color(black)("mol CO")))) = "0.750 mol CO}}_{2}$

${\text{Moles of Na"_2"CO"_3color(white)(l) "from CO" = 0.750 color(red)(cancel(color(black)("mol CO"_2))) × ("1 mol Na"_2"CO"_3)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "0.750 mol Na"_2"CO}}_{3}$

$\text{Moles of NaOH from CO" = 0.750 color(red)(cancel(color(black)("mol Na"_2"CO"_3))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol Na"_2"CO"_3)))) = "1.50 mol NaOH}$

$\text{Total moles NaOH" = "moles from CO + moles from CO"_2 = "1.50 mol + 0.500 mol = 2.00 mol}$

$\text{Mass of NaOH" = 2.00 color(red)(cancel(color(black)("mol NaOH"))) × "40.00 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "80 g NaOH}$

Jun 18, 2017

Here's what I got.

Explanation:

The idea here is that a solution of carbon dioxide is acidic because of the formation of carbonic acid, ${\text{H"_2"CO}}_{3}$

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

The sodium hydroxide will then neutralize the acid and produce aqueous sodium carbonate and water

${\text{H"_ 2"CO"_ (3(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

Since you know that carbonic acid is best represented as the equilibrium that involves aqueous carbon dioxide and water, you can say that you have

overbrace("CO"_ (2(aq)) + color(red)(cancel(color(black)("H"_ 2"O"_ ((l))))))^(color(blue)("H"_ 2"CO"_ (3(aq)))) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + color(red)(cancel(color(black)(2)))"H"_ 2"O"_ ((l))

which is equivalent to

${\text{CO"_ (2(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O}}_{\left(l\right)}$

Now, notice that every mole of aqueous carbon dioxide consumes $2$ moles of sodium hydroxide.

Use the molar mass of the sodium hydroxide to convert the first sample to moles

20 color(red)(cancel(color(black)("g"))) * "1 mole NaOH"/(40color(red)(cancel(color(black)("g")))) = "0.5 moles NaOH"

This means that the mixture of carbon monoxide and carbon dioxide contained

0.5 color(red)(cancel(color(black)("moles NaOH"))) * "1 mole CO"_2/(2color(red)(cancel(color(black)("moles NaOH")))) = "0.25 moles CO"_2

and

overbrace("1 mole")^(color(blue)("CO"_ ((g)) + "CO"_ (2(g)))) - "0.25 moles CO"_2 = "0.75 moles CO"

Now, when you're oxidizing the carbon monoxide to carbon dioxide, you're essentially converting it to carbon dioxide.

${\text{0.75 moles CO " stackrel(color(white)(acolor(blue)("oxidation")aaa))(->) "0.75 moles CO}}_{2}$

This implies that the $1$ mole of the mixture is now $1$ mole of carbon dioxide.

The reaction will now require

1 color(red)(cancel(color(black)("mole CO"_2))) * "2 moles NaOH"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "2 moles NaOH"

which are equivalent to

2 color(red)(cancel(color(black)("moles NaOH"))) * "40 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = "80 g"#

So, you know that you need $\text{20 g}$ of sodium hydroxide to convert $0.25$ moles of carbon dioxide to sodium carbonate and $\text{60 g}$ of sodium hydroxide to convert $0.75$ moles of carbon dioxide to sodium carbonate, hence why you'd need $\text{80 g}$ of sodium hydroxide to convert $1$ mole of carbon dioxide to sodium carbonate.