# Question #de715

May 19, 2017

The mass of excess reactant remaining is 272 g.

#### Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation, masses, and moles, so let's gather all the information in one place.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m m m m} 17.03 \textcolor{w h i t e}{m m l} 46.01$
$\textcolor{w h i t e}{m m m m m m} \text{6NH"_3 + color(white)(ll)"8NO"_2 → "7N"_2"O" + "9H"_2"O}$
$\text{Mass/g:} \textcolor{w h i t e}{m m l l} 352 \textcolor{w h i t e}{m m m} 289$
$\text{Moles:} \textcolor{w h i t e}{m m m} 20.67 \textcolor{w h i t e}{m m} 6.281$
$\text{Divide by:} \textcolor{w h i t e}{m m} 6 \textcolor{w h i t e}{m m m m l} 8$
$\text{Moles rxn:} \textcolor{w h i t e}{m} 3.445 \textcolor{w h i t e}{m l l} 0.7852$

${\text{Moles of NH"_3 = 352 color(red)(cancel(color(black)("g NH"_3))) × "1 mol NH"_3/(17.03 color(red)(cancel(color(black)("g NH"_3)))) = "20.67 mol NH}}_{3}$

${\text{Moles of NO"_2 = 289 color(red)(cancel(color(black)("g NO"_2))) × ("1 mol NO"_2)/(46.01 color(red)(cancel(color(black)("g SO"_2)))) = "6.281 mol NO}}_{2}$

a) Identify the limiting reactant

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

You simply divide the number of moles by the corresponding coefficient in the balanced equation.

I did that for you in the chart above.

${\text{NO}}_{2}$ is the limiting reactant because it gives the fewest "moles of reaction".

${\text{NH}}_{3}$ is then the excess reactant.

b) Calculate the mass of ${\text{NH}}_{3}$ reacted

${\text{Mass of NH"_3 = 6.281 color(red)(cancel(color(black)("mol NO"_2))) × (6 color(red)(cancel(color(black)("mol NH"_3))))/(8color(red)(cancel(color(black)("mol NO"_2)))) × ("17.03 g NH"_3)/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "80.22 g NH}}_{3}$

c) Calculate the mass of ${\text{NH}}_{3}$ remaining

$\text{Mass of NH"_3 = "352 g - 80.22 g = 272 g}$