Question #a20b7

Start with the balanced equation for the reaction

#M_text(r):color(white)(m)40.00#
#color(white)(mmll)"2NaOH" + "CO"_2 → "Na"_2"CO"_3 + "H"_2"O"#

Calculate the moles of #"NaOH"#

#"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 #"CO"_2# in the mixture

#" 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 #"CO"# in the mixture

#"Moles of CO + Moles of CO"_2 = "total moles"#

#"Moles of CO + 0.250 mol = 1 mol"#

#"Moles of CO = (1 - 0.250) mol = 0.750 mol"#

Write the balanced equation for the oxidation of #"CO"#

#"2CO + O"_2 → "2CO"_2#

#"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#

#"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#

#"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"#

#"Total moles NaOH" = "moles from CO + moles from CO"_2 = "1.50 mol + 0.500 mol = 2.00 mol"#

#"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"#

The idea here is that a solution of carbon dioxide is acidic because of the formation of carbonic acid, #"H"_2"CO"_3#

#"CO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 2"CO"_ (3(aq))#

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

#"H"_ 2"CO"_ (3(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + 2"H"_ 2"O"_ ((l))#

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

#"CO"_ (2(aq)) + 2"NaOH"_ ((aq)) -> "Na"_ 2"CO"_ (3(aq)) + "H"_ 2"O"_ ((l))#

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.

#"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 #"20 g"# of sodium hydroxide to convert #0.25# moles of carbon dioxide to sodium carbonate and #"60 g"# of sodium hydroxide to convert #0.75# moles of carbon dioxide to sodium carbonate, hence why you'd need #"80 g"# of sodium hydroxide to convert #1# mole of carbon dioxide to sodium carbonate.