# Question 890a6

Jun 11, 2017

$\text{12 g C}$

#### Explanation:

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

Start by writing the balanced chemical equation that describes this reaction

${\text{C"_ ((s)) + "O"_ (2(g)) -> "CO}}_{2 \left(g\right)}$

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

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{unreacted carbon = 12 g}}}}$

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.