# Question d20e1

Sep 3, 2017

WARNING! Long answer! The limiting reactant is ${\text{O}}_{2}$; theoretical yield = 5,97 g; percent yield = 600 %.

#### Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and everything else below them.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m} \text{28,02"color(white)(mll)"32,00"color(white)(mml)"44,01}$
$\textcolor{w h i t e}{m m m m m l} {\text{2CO" +color(white)(m) "O"_2 → color(white)(m)2"CO}}_{2}$
$\text{Mass/g:"color(white)(mm)25color(white)(mmll)"2,17} \textcolor{w h i t e}{m m m l} 36$
$\text{Moles:" color(white)(mmml)"0,892"color(white)(ll)"0,067 81} \textcolor{w h i t e}{m l l}$
$\text{Divide by:} \textcolor{w h i t e}{m l} 2 \textcolor{w h i t e}{m m m} 1$
$\text{Moles rxn:"color(white)(ml)"0,446"color(white)(m)"0,067 81}$

(a) Calculate the moles of $\text{CO}$

$\text{Moles of CO" = 25 color(red)(cancel(color(black)("g CO"))) × ("1 mol CO")/("28,02" color(red)(cancel(color(black)("g CO")))) = "1,12 mol CO}$

(b) Calculate moles of ${\text{O}}_{2}$

${\text{2,17"color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/("32,00" color(red)(cancel(color(black)("g O"_2)))) = "0,067 81 mol O}}_{2}$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

${\text{O}}_{2}$ is the limiting reactant because it gives the fewer moles of reaction.

3. Calculate the theoretical yield of ${\text{CO}}_{2}$.

${\text{Theoretical yield" = "0,067 81" color(red)(cancel(color(black)("mol O"_2))) × (2 color(red)(cancel(color(black)("mol CO"_2))))/(1 color(red)(cancel(color(black)("mol O"_2)))) × "44,01 g CO"_2/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "5,97 g CO}}_{2}$

The theoretical yield of ${\text{CO}}_{2}$ is 5,97 g.

4. Calculate the percent yield of ${\text{CO}}_{2}$

The formula for percentage yield is

color(blue)(bar(ul(|color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "

"Percent yield" = (36 color(red)(cancel(color(black)("g"))))/("5,97" color(red)(cancel(color(black)("g")))) × 100 % = 600 %#

The percent yield is 600 %.

This is an impossible result. Are you sure you gave us the correct data?