# How many grams of carbon dioxide are produced when "83.7 g" of carbon monoxide react with excess iron(II) oxide?

Apr 26, 2016

${\text{132 g CO}}_{2}$

#### Explanation:

Start by taking a look at the balanced chemical equation that describes this redox reaction

${\text{Fe"_ 2"O"_ (3(s)) + 3"CO"_ ((g)) -> 2"Fe"_ ((s)) + 3"CO}}_{2 \left(g\right)}$

According to the chemical equation, $3$ moles of carbon monoxide, $\text{CO}$, will react with $1$ mole of ferric oxide, ${\text{Fe"_2"O}}_{3}$, and produce $3$ moles of carbon dioxide, ${\text{CO}}_{2}$.

Now, you are given grams of carbon monoxide and asked for grams of carbon dioxide. This means that you'll have to convert the aforementioned mole ratio that exists between carbon monoxide and carbon dioxide to a gram ratio.

Since no mention of the mass of ferric oxide was made, you can assume that this reactant is in excess.

So, to convert between moles and grams, use the molar mass of the two compounds.

${\text{For CO: " M_M = "28.01 g mol}}^{- 1}$

${\text{For CO"_2: color(white)(a)M_M = "44.01 g mol}}^{- 1}$

If one mole of carbon monoxide has a mass of $\text{28.01 g}$ and one mole of carbon dioxide has a mass of $\text{44.01 g}$, it follows that the reaction produces $\text{44.01 g}$ of carbon dioxide for every $\text{28.01 g}$ of carbon monoxide.

This is the case because of the $3 : 3$ mole ratio that exists between the two compounds.

So, your $\text{83.7 g}$ of carbon monoxide will produce

$83.7 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{g CO"))) * "44.01 g CO"_2/(28.01 color(red)(cancel(color(black)("g CO")))) = color(green)(|bar(ul(color(white)(a/a)"132 g CO}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.