# Question #5bb41

Jul 27, 2017

${\text{87 g O}}_{2}$

#### 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

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

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 $\text{44.01 g}$ and $1$ mole of oxygen gas has a mass of $\text{32.0 g}$--look up the molar masses of the two chemical species--you can say that for every $\text{32.0 g}$ of oxygen gas consumed in the reaction, you get $\text{44.01 g}$ of carbon dioxide.

This implies that in order to produce $\text{120 g}$ of carbon dioxide, the reaction needs to consume

$120 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g CO"_2))) * overbrace("32.0 g O"_2/(44.01color(red)(cancel(color(black)("g CO"_2)))))^(color(blue)("the equivalent of a 1:1 mole ratio")) = color(darkgreen)(ul(color(black)("87 g O}}_{2}}}}$

The answer is rounded to two sig figs.