# Question #9b1cc

##### 1 Answer

#### Explanation:

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

Once you know that, use an **ICE table** to find the equilibrium concentration of carbon dioxide, *grams* of carbon dioxide.

So, heating carbon monoxide to

As you can see from the value of the *equilibrium constant*, **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 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))#

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 >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)|)))#