Question #5bb41

1 Answer
Jul 27, 2017

Answer:

#"87 g O"_2#

Explanation:

For starters, you need a balanced chemical equation to go by. Since you didn't provide one, I'll assume that this is the reaction you're working with

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

Notice that for every #1# mole of oxygen gas that the reaction consumes, you get #1# mole of carbon dioxide.

Since you know that #1# mole of carbon dioxide has a mass of #"44.01 g"# and #1# mole of oxygen gas has a mass of #"32.0 g"#--look up the molar masses of the two chemical species--you can say that for every #"32.0 g"# of oxygen gas consumed in the reaction, you get #"44.01 g"# of carbon dioxide.

This implies that in order to produce #"120 g"# of carbon dioxide, the reaction needs to consume

#120 color(red)(cancel(color(black)("g CO"_2))) * overbrace("32.0 g O"_2/(44.01color(red)(cancel(color(black)("g CO"_2)))))^(color(blue)("the equivalent of a 1:1 mole ratio")) = color(darkgreen)(ul(color(black)("87 g O"_2)))#

The answer is rounded to two sig figs.