Question

Question #5bb41

Answer 1

`"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.