Question #890a6

I am going to assume that you're working with #"20 g"# of carbon and #"20 g"# of oxygen gas, not with #"20 g"# of carbon and oxygen gas, i.e. that the mass of carbon and the mass of oxygen gas add up to give #"40 g"#, not #"20 g"#.

Start by writing the balanced chemical equation that describes this reaction

#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#

Notice that the reaction consumes #1# mole of oxygen gas for every #1# moles of carbon that takes part in the reaction.

Convert the two masses to moles by using the molar mass of carbon and the molar mass of oxygen gas, respectively.

#20 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "1.665 moles C"#

#20 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.625 moles O"_2#

At this point, it should be clear that oxygen gas acts as a limiting reagent, i.e. it gets completely consumed before all the moles of carbon get the chance to react.

This is the case because #1.665# moles of carbon would require #1.665# moles of oxygen gas to react completely. Since you only have #0.625# moles of oxygen gas available, you can say that the reaction will consume #0.625# moles of oxygen gas and of carbon and leave you with

#overbrace("1.665 moles C")^(color(blue)("what is available")) - overbrace("0.625 moles C")^(color(blue)("what is consumed")) = overbrace("1.040 moles C")^(color(blue)("what remains unreacted"))#

To convert this to grams, use the molar mass of carbon

#1.040 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = "12.49 g"#

Therefore, you can say that after the reaction is complete, you will be left with

#color(darkgreen)(ul(color(black)("unreacted carbon = 12 g")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the two masses.