# Question #5b51f

Jan 24, 2018

Here's one way to do it.

#### Explanation:

Problem:

Assume you are reacting 20.0 g of carbon and 20.0 g of oxygen to form carbon dioxide. What is the mass of excess reactant used and what mass of excess reactant is unused?

Solution:

${M}_{r} : \textcolor{w h i t e}{m m m l l} 12.01 \textcolor{w h i t e}{m m} 32.00$
$\textcolor{w h i t e}{m m m m m m m} {\text{C" color(white)(m)+ color(white)(m)"O"_2color(white)(m) →color(white)(m) "CO}}_{2}$
$\text{Mass/g:} \textcolor{w h i t e}{m m m} 20 \textcolor{w h i t e}{m m m l} 20$
$\text{Amt/mol:} \textcolor{w h i t e}{m l l} 1.67 \textcolor{w h i t e}{m m} 0.625$

You calculated that:

• The amount of ${\text{O}}_{2}$ is 0.625 mol
• The amount of $\text{C}$ is 1.67 mol
• ${\text{O}}_{2}$ is the limiting reactant
• $\text{C}$ is the excess reactant

Then

$\text{Moles of C used" = 0.625 color(red)(cancel(color(black)("mol O"_2))) × "1 mol C"/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.625 mol C}$

$\text{Mass of C used" = 0.625 color(red)(cancel(color(black)("mol C"))) × "12.01 g C"/(1 color(red)(cancel(color(black)("mol C")))) = "12.5 g C}$

$\text{Mass of unused C = 20.0 g - 12.5 g = 7.5 g}$

Therefore,

• Mass of excess reactant used = 12.5 g
• Mass of unused excess reactant = 7.5 g