Question #5c516
1 Answer
Explanation:
Start by writing the balanced chemical equation that describes your reaction
#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#
Notice that it takes
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
Oxygen gas,
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 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.