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

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

Answer:
\(132 \mathrm{~g} \mathrm{CO}_{2}\)

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

\(\mathrm{Fe}_{2} \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)}\)

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

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.

If one mole of carbon monoxide has a mass of 28.01 g and one mole of carbon dioxide has a mass of 44.01 g, it follows that the reaction produces 44.01 g of carbon dioxide for every 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 83.7 g of carbon monoxide will produce

The answer is rounded to three sig figs.