A mixture of 0.166 mols of #"C"# reacted with 0.117 mols of #"O"_2# in a sealed, 10.0 L vessel at 500.0 K, producing a mixture of #"CO"# and #"CO"_2#. The reaction is:

#3"C"(s)+2"O"_2(g)- 2"CO"(g)+"CO"_2(g)#. What is the partial pressure of #"CO"#?

Question

The total pressure is 0.707 atm.

Peyton Adams · Accepted Answer

Here's what I got.

The balanced chemical equation that describes this reaction #3"C"_ ((s)) + 2"O"_ (2(g)) -> 2"CO"_ ((g)) + "CO"_ (2(g))# tells you that for every #3# moles of carbon that take part in the reaction, the reaction consumes #2# moles of oxygen gas. Use this #3:2# mole ratio to determine if one of the two reactants will act as a limiting reagent. In order for all the moles of carbon to react, you would need #0.166 color(red)(cancel(color(black)("moles C"))) * "2 moles O"_2/(3color(red)(cancel(color(black)("moles C")))) = "0.1107 moles O"_2# Since your sample contains more moles than you would need to ensure that all the moles of carbon take part in the reaction, you can say that oxygen gas will be in excess, i.e. carbon will act as the limiting reagent. The reaction will consume #0.166# moles of carbon and #0.1107# moles of oxygen gas and leave you with #"0.117 moles O"_2 - "0.1107 moles O"_2 = "0.0063 moles O"_2" "# The reaction will produce #0.166 color(red)(cancel(color(black)("moles C"))) * "2 moles CO"/(3color(red)(cancel(color(black)("moles C")))) = "0.1107 moles CO"# and #0.166 color(red)(cancel(color(black)("moles C"))) * "1 mole CO"_2/(3color(red)(cancel(color(black)("moles C")))) = "0.05533 moles CO"_2# Now, in order to find the partial pressure of carbon monoxide, you need to first figure out its mole fraction in the mixture. The mole fraction of carbon monoxide, #chi_"CO"#, is defined as the ratio between the number of moles of carbon monoxide and the total number of moles of gas present in the mixture. In your case, the mixture will contain #overbrace("0.0063 moles O"_2)^(color(blue)("excess O"_2)) + "0.1107 moles CO" + "0.05533 moles CO"_2 = "0.1723 moles gas"# This means that you have #chi_ "CO" = (0.1107 color(red)(cancel(color(black)("moles CO"))))/(0.1723color(red)(cancel(color(black)("moles gas")))) = 0.6425# Finally, to find the partial pressure of carbon monoxide, use Dalton's Law of Partial Pressures, which states that the partial pressure of a gas that's part of a gaseous mixture depends on the mole fraction of said gas and on the total pressure of the mixture. #P_ "CO" = chi_ "CO" * P_"total"# In your case, you have #P_ "CO" = 0.6425 * "0.707 atm"# #color(darkgreen)(ul(color(black)(chi_ "CO" = "0.454 atm")))# I'll leave the answer rounded to three sig figs.

Wyatt Adkins · Answer

undefined The partial pressure of "CO" is 0.100 atm.

A mixture of 0.166 mols of #"C"# reacted with 0.117 mols of #"O"_2# in a sealed, 10.0 L vessel at 500.0 K, producing a mixture of #"CO"# and #"CO"_2#. The reaction is: #3"C"(s)+2"O"_2​(g)-&gt; 2"CO"(g)+"CO"_2​(g)#. What is the partial pressure of #"CO"#?

The total pressure is 0.707 atm.

A mixture of 0.166 mols of #"C"# reacted with 0.117 mols of #"O"_2# in a sealed, 10.0 L vessel at 500.0 K, producing a mixture of #"CO"# and #"CO"_2#. The reaction is: #3"C"(s)+2"O"_2(g)-> 2"CO"(g)+"CO"_2(g)#. What is the partial pressure of #"CO"#?