Question #9b1cc

Your strategy here will be to use the volume of the reaction vessel and the number of moles of carbon monoxide, #"CO"#, to calculate the initial concentration of the reactant.

Once you know that, use an ICE table to find the equilibrium concentration of carbon dioxide, #"CO"_2#, then convert this to grams of carbon dioxide.

So, heating carbon monoxide to #850^@"C"# will cause an equilibrium reaction that forms carbon dioxide, #"CO"_2#< and carbon, #"C"#, to be established.

As you can see from the value of the equilibrium constant, #K_c >1#, this equilibrium will lie mostly to the right, i.e. the forward reaction will be favored.

You can thus expect most of the carbon monoxide to be converted to carbon dioxide and carbon at equilibrium.

Use the molar mass of carbon monoxide to find the number of moles in that #"21-g"# sample

#21 color(red)(cancel(color(black)("g"))) * "1 mole CO"/(28.01color(red)(cancel(color(black)("g")))) = "0.750 moles CO"#

The initial concentration of the reactant will be

#["CO"]_0 ="0.750 moles"/"2.5 L" = "0.30 mol L"^(-1)#

Set up your ICE table to find the equilibrium concentration of carbon dioxide - keep in mind that the concentration of the solid is considered constant!

#" " color(red)(2)"CO"_ ((g)) " "rightleftharpoons" " "CO"_ (2(g)) " "+" " "C"_ ((s))#

#color(purple)("I")color(white)(aaaaacolor(black)(0.30)aaaaaaaaaaacolor(black)(0)aaaaaaaaaacolor(black)(-)#
#color(purple)("C")color(white)(aaacolor(black)((-color(red)(2)x))aaaaaaaacolor(black)((+x))aaaaaaaacolor(black)(-)#
#color(purple)("E")color(white)(aacolor(black)(0.30-color(red)(2)x)aaaaaaaaacolor(black)(x)aaaaaaaaaacolor(black)(-)#

By definition, the equilibrium constant will be equal to

#K_c = (["CO"_2])/(["CO"]^color(red)(2))#

In your case, this will be equal to

#10.0 = x/(0.30 - color(red)(2)x)^color(red)(2)#

Rearrange to get

#10.0 * (0.09 - 1.2x + 4x^2) = x#

#39x^2 - 12x + 0.9 = 0#

This quadratic equation will produce two positive solutions

#color(red)(cancel(color(black)(x_1 = 0.1782)))" "# and #" "x_2 = 0.1295color(white)(a)color(green)(sqrt())#

Notice that the equilibrium concentration of carbon monoxide is equal to #(0.30 - 2x)#, which means that the valid solution will have to satisfy

#0.30 - 2x >0#

This is why the valid solution is

#x = 0.1295#

The equilibrium concentration of carbon dioxide will be

#["CO"_2] = "0.1295 mol L"^(-1)#

Use the volume of the reaction vessel to convert this to moles of carbon dioxide

#2.5 color(red)(cancel(color(black)("L"))) * "0.1295 moles"/(1color(red)(cancel(color(black)("L")))) = "0.32375 moles CO"_2#

Finally, use the molar mass of carbon dioxide to convert the moles to grams

#0.32375color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "14.25 g"#

Rounded to two sig figs, the number of sig figs you have for the volume of the reaction vessel and the mass of carbon monoxide, the answer will be

#"mass of CO"_2 = color(green)(|bar(ul(color(white)(a/a)"14 g"color(white)(a/a)|)))#