Question #5c516

1 Answer
Mar 18, 2017

Answer:

#"22 g CO"_2#

Explanation:

Start by writing the balanced chemical equation that describes your reaction

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

Notice that it takes #1# mole of oxygen gas in order to react with #1# mole of carbon and produce #1# mole of carbon dioxide.

Now, your goal here is to figure out if you have enough oxygen gas to ensure that all the moles of carbon take part in the reaction.

Since you have #2# moles of carbon available, you must check to see if the sample of oxygen gas contains at least #2# moles of this reactant.

Oxygen gas, #"O"_2#, has a molar mass of #"32.0 g mol"^(-1)#, which means that #1# mole of oxygen gas has a mass of #"32.0 g"#.

Your sample of oxygen gas will thus contain

#16 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.50 moles O"_2#

This means that oxygen gas acts as a limiting reagent because it gets completely consumed before all the moles of carbon get the chance to react.

You can say that the reaction will consume #0.50# moles of oxygen gas and

#0.50 color(red)(cancel(color(black)("moles O"_2))) * "1 mole C"/(1color(red)(cancel(color(black)("mole O"_2)))) = "0.50 moles C"#

and produce

#0.50 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole O"_2)))) = "0.50 moles CO"_2#

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

#0.50 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(darkgreen)(ul(color(black)("22 g")))#

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