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.