# Question 6f862

Jun 27, 2017

$47$ ${\text{g CO}}_{2}$

#### Explanation:

We're asked to find the mass of ${\text{CO}}_{2}$ that forms when $30$ $\text{g CO}$ are burned in excess oxygen.

The equation for this reaction is

$2 {\text{CO" (g) + "O"_2(g) rarr 2"CO}}_{2} \left(g\right)$

Let's first calculate the number of moles of carbon monoxide, using its molar mass, $28.01$ $\text{g/mol}$:

30cancel("g CO")((1color(white)(l)"mol CO")/(28.01cancel("g CO"))) = color(red)(1.07 color(red)("mol CO"

Since $\text{CO}$ and ${\text{CO}}_{2}$ have the same coefficients, the relative number of moles will be the same:

color(red)(1.07)color(white)(l)color(red)("mol CO") = color(green)(1.07)color(white)(l)color(green)("mol CO"_2

Lastly, let's use the molar mass of ${\text{CO}}_{2}$ ($44.01$ $\text{g/mol}$) to calculate the theoretical mass yield of carbon dioxide:

color(green)(1.07)cancel(color(green)("mol CO"_2))((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(47 color(blue)("g CO"_2

rounded to $2$ significant figures.

Thus, for this reaction, if $30$ grams of carbon monoxide are burned in air, color(blue)(47 sfcolor(blue)("grams of CO"_2# can form (this is the theoretical yield, the maximum amount that can form).