# Question 93107

Apr 18, 2017

The theoretical yield is 53.8 g of ${\text{CO}}_{2}$. The percent yield is 97.9 %.

#### Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation, molar masses, moles, etc, so let's gather them all in one place now.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m l} 16.04 \textcolor{w h i t e}{m l} 32.00 \textcolor{w h i t e}{m l} 44.01$
$\textcolor{w h i t e}{m m m m m m} \text{CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O}$
$\text{Mass/g:} \textcolor{w h i t e}{m m} 23.2 \textcolor{w h i t e}{m l l} 78.3$
$\text{Moles:} \textcolor{w h i t e}{m m l} 1.446 \textcolor{w h i t e}{m l} 2.447$
$\text{Divide by:} \textcolor{w h i t e}{m l l} 1 \textcolor{w h i t e}{m m m} 2$
$\text{Moles rxn:} \textcolor{w h i t e}{l} 1.446 \textcolor{w h i t e}{m l} 1.223$

Identify the limiting reactant

A convenient way to identify the limiting reactant is to calculate its "moles of reaction".

You divide its number of moles by the coefficient in the balanced equation. I did that for you above.

We see that ${\text{O}}_{2}$ is the limiting reactant because it gives the fewest moles of reaction.

Calculate the theoretical yield

${\text{Theor. yield" = 2.447 color(red)(cancel(color(black)("mol O"_2))) × (1 color(red)(cancel(color(black)("mol CO"_2))))/(2 color(red)(cancel(color(black)("mol O"_2)))) × ("44.01 g CO"_2)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "53.8 g CO}}_{2}$

Calculate the % yield

"Percent yield" = (52.7 color(red)(cancel(color(black)("g"))))/(53.84 color(red)(cancel(color(black)("g")))) ×100 % = 97.9 %#