# Question ddfaa

Feb 8, 2016

"% yield" = 80.%

#### Explanation:

For a given reaction, the theoretical yield tells you the maximum amount of product that can formed following that reaction.

In your case, the balanced chemical equation for the combustion of propane, ${\text{C"_3"H}}_{8}$, looks like this

${\text{C"_3"H"_text(8(g]) + color(red)(5)"O"_text(2(g]) -> color(blue)(3)"CO"_text(2(g]) + color(green)(4)"H"_2"O}}_{\textrm{\left(l\right]}}$

This tells you that for every $1$ mole of propane that undergoes decomposition, the reaction will always consume

• $\textcolor{red}{5}$ moles of oxygen gas

and theoretically produce

• $\textcolor{b l u e}{3}$ moles of carbon dioxide
• $\textcolor{g r e e n}{4}$ moles of water

Now, the theoretical yield of the reaction is what you get for a 100% yield.

Simply put, if every single molecule of propane reacts with five molecules of oxygen and produces three molecules of carbon dioxide and four molecules of water, then the reaction is said to have a 100% yield.

Before doing anything else, you must figure out if one of the two reactants acts as a limiting reagent. To do that, use the molar mass of propane to determine how many moles you have present

33 color(red)(cancel(color(black)("g"))) * ("1 mole C"_3"H"_8)/(44.1color(red)(cancel(color(black)("g")))) = "0.7483 moles C"_3"H"_8

Use the molar volume of a gas at STP to find the number of moles of oxygen gas. I will use the current definition of STP, which implies a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions, one mole of any ideal gas occupies $\text{22.7 L}$, not $\text{22.4 L}$. If your instructor wants you to use the old value, redo the calculations using $\text{22.4 L}$.

56 color(red)(cancel(color(black)("L O"_2))) * overbrace("1 mole O"_2/(22.7color(red)(cancel(color(black)("L O"_2)))))^(color(purple)("molar volume of a gas")) = "2.3467 moles O"_2

So, do you have enough oxygen to ensure that all the moles of propane react?

In order for all the moles of propane to react, you'd need

0.7483 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(red)(5)" moles O"_2)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "3.742 moles O"_2

Since you don't have enough oxygen, oxygen gas will act as a limiting reagent, i.e it will determine how much propane actually undergoes combustion.

More specifically, this much oxygen would allow for

2.3467 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole C"_3"H"_8)/(color(red)(5)color(red)(cancel(color(black)("moles O"_2)))) = "0.4693 moles C"_3"H"_8

to take part of the reaction. The rest will remain in excess.

Now focus on finding the theoretical yield of the reaction. If $0.4693$ moles of propane react, how many moles of water should be produced for a 100% yield?

0.4693 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(green)(4)" moles H"_2"O")/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "1.877 moles H"_2"O"

To get the mass of water, use the compound's molar mass

1.877 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "33.81 g"

However, you are told that the reaction produces $\text{27 g}$ of water. This means that it does not have a 100% yield. As you know, percent yield is defined as

$\textcolor{b l u e}{\text{% yield" = "actual yield"/"theoretical yield} \times 100}$

In your case, the theoretical yield is $\text{33.81 g}$, and the actual yield is $\text{27 g}$. This means that the reaction's percent yield is equal to

"% yield" = (27 color(red)(cancel(color(black)("g"))))/(33.81color(red)(cancel(color(black)("g")))) xx 100 = color(green)("80. %") -># rounded to two sig figs