# Question 2d295

May 21, 2017

$\text{0.878 g CO}$

#### Explanation:

Start by looking at the balanced chemical equation that describes this reaction

$\textcolor{b l u e}{2} {\text{CO"_ ((g)) + "O"_ (2(g)) -> 2"CO}}_{2 \left(g\right)}$

You know that every $1$ mole of oxygen gas that takes part in the reaction consumes $\textcolor{b l u e}{2}$ moles of carbon monoxide and produces $2$ moles of carbon dioxide.

You already know the mass of carbon dioxide produced by the reaction, so use the compound's molar mass to convert it to moles

1.38 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01 color(red)(cancel(color(black)("g")))) = "0.03136 moles CO"_2

Now, you can assume that the oxygen gas is in excess because the problem doesn't mention a specific sample of this reactant.

Use the $\textcolor{b l u e}{2} : 2$ mole ratio that exists between carbon monoxide and carbon dioxide to find the number of moles of the former needed to produce that much product

0.03136 color(red)(cancel(color(black)("moles CO"_2))) * (color(blue)(2)color(white)(.)"moles CO")/(2color(red)(cancel(color(black)("moles CO"_2)))) = "0.03136 moles CO"#

Finally, convert this to grams by using the compound's molar mass

$0.03136 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CO"))) * "28.01 g"/(1color(red)(cancel(color(black)("mole CO")))) = color(darkgreen)(ul(color(black)("0.878 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of carbon dioxide.