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AP Chemistry : Stoichiometry

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Example Question #11 : Balancing Equations

Transesterification is an industrially important process for the production of biodiesel fuel (that most diesel engines can run cleanly on) and glycerin from triglycerides (usually people use vegetable oil as their feedstock).

One easy to do and high yielding method involves the reaction of ethanol or methanol with sodium hydroxide to produce the super bases sodium ethoxide or sodium methoxide respectively. The super base is then used in the transesterification reaction with triglycerides to produce glycerin and ethyl or methyl esters (biodiesel). The products can easily be separated by differences in density. Note: For this reaction to occur the reactants must be free of water as when water is present saponification occurs (how soap is made) which will compete with the transesterification reaction. Note: just because the reaction described above is easy doesn't mean that it is anywhere near safe; please do thoroughly research any reaction you plan to undertake and take all possible safety precautions. That being said, this process is becoming much more popular lately and with the recent focus on alternative energy may well soon account for a decent portion of fuel production.

Catalyst formation equation:

$NaOH+CH_3OH\rightleftharpoons Na^+CH_3OH+H_2O$

Unbalanced transesterification equation:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

Balance the following transesterification reaction assuming that all R-groups are identical:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

Possible Answers:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

$2RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+4CH_3CO_2R$

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

$3RCO_2CH_2CH(O_2CR)CH_2CO_2R +2CH_3OH \xrightarrow[]{Catalyst} 5C_3H_8O_3+CH_3CO_2R$

$\frac{1}{2}RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+2CH_3CO_2R$

Correct answer:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

Explanation:

The steps for balancing equations are as follows and should be done in order:

1.) Check for diatomic molecules: In this case there are none so move on to step 2

2.) Balance the metals (hydrogen is not considered a metal in this application): We have no metals in our equation, so move on to step 3 (the sodium is part of the catalyst so we don't consider it part of the balancing).

3.) Balance the nonmetals (not including oxygen). If you notice the left hand side of the equation has 3R groups and the right hand side only has one R-group. To fix this we can put a coefficient of 3 in front of CH_3CO_2R on the right hand side giving us the following:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

On the left hand side of the equation we have 7 carbon atoms but on the right hand side we have 9 carbon atoms. This can be rectified by putting a coefficient of 3 in front of CH_3OH on the left hand side of the equation giving us the following:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

Now all the nonmetals (excluding oxygen) have been balanced and we can move on to step 4

4.) Balance oxygen. There are 9 oxygen atoms on both sides of the equation. Move on to step 5

5.) Balance Hydrogen. There are 17 hydrogens on both sides of the equation.

6.) Recount all atoms to make sure you have balanced correctly. 9 carbons on both sides, 3 R-groups on both sides, 9 oxygen on both sides, and 17 hydrogens on both sides.

If this problem had involved ionic species you would also want to make certain that the charges are balanced on both sides of the equation.

The charges on both sides add up to zero so your equation is balanced:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

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Example Question #431 : Ap Chemistry

Consider the following unbalanced reaction:

$AgNO_{3}+K_{3}PO_{4}\rightarrow Ag_{3}PO_{4}+KNO_{3}$

Once this equation has been balanced, what are the the respective stoichiometric coefficients (listed in the order in which the above compounds appear in the reaction)?

Possible Answers:

$1,\:3,\:3,\:1$

$1,\:3,\:1,\:3$

$3,\:1,\:3,\:1$

$3,\:1,\:1,\:3$

Correct answer:

$3,\:1,\:1,\:3$

Explanation:

To balance chemical reactions, it's usually more effective to save the more common elements, such as oxygen, for last. For this reaction, let's go ahead and start by balancing the element on both sides.

$3AgNO_{3}+K_{3}PO_{4}\rightarrow Ag_{3}PO_{4}+KNO_{3}$

Because there are moles of on the right side, we'll need to have moles of $AgNO_{3}$ . Next, let's go ahead and balance nitrogen on both sides.

$3AgNO_{3}+K_{3}PO_{4}\rightarrow Ag_{3}PO_{4}+3KNO_{3}$

Since we added a coefficient of to the $AgNO_{3}$ in the first step, there are now moles of on the left. Thus, we'll need to add a coefficient of to the $KNO_{3}$ . Next, if we check the amount of , , and on both sides, we see that they are equal. Thus, we can recognize that $K_{3}PO_{4}$ and $Ag_{3}PO_{4}$ have coefficients of , and do not need to be altered.

Our final balanced equation looks like this.

$3AgNO_{3}+1K_{3}PO_{4}\rightarrow 1Ag_{3}PO_{4}+3KNO_{3}$

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Example Question #1 : Limiting Reagent

Consider the following balanced reaction:

2Ca(s) + O₂(g) → 2CaO

Consider the reaction above. If 60g of Ca is placed in a reaction chamber with 32g of O₂ which will be the limiting reactant?

Possible Answers:

O₂ because it is a diatomic molecule

Ca because there are 1.5 moles of Ca and only 1 mole of O₂

O₂ because there is less of it

There will not be a limiting reactant

Ca because there will need to be 2 moles if Ca for every mole of O₂ as defined by the balanced equation.

Correct answer:

Ca because there will need to be 2 moles if Ca for every mole of O₂ as defined by the balanced equation.

Explanation:

When considering Limiting Reactant problems the most important aspect to consider is the molar ratio of the reactants. Here the balanced formula tells us that for every 2 moles of Ca there must be 1 mole of O₂ to create the product. The amounts given by the problem are the actual amounts we are given and can be compared to the molar ratio to determine the limiting reactant. Since there is 1 mole of O_2,in any situation in which there are less than 2 moles of Ca, Ca is the limting reactant. Conversely if there were 2 moles of Ca, any situation in which there less than 1 mole of O₂ would be a situation in which O₂ is the limiting reactant.

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Example Question #1 : Limiting Reagent

Given the reaction

H₂SO₄ (aq) + 2 LiOH (aq) → Li₂SO₄ (aq) + 2 H₂O (l)

If you have 100g of H2SO4 and 65g of LiOH. What is your limiting reactant?

Possible Answers:

LiOH

Both H₂SO₄ and LiOH limit the reaction equally

H₂O

Li₂SO₄

H₂SO₄

Correct answer:

H₂SO₄

Explanation:

H₂SO₄ molecular weight is 98. This gives 1.02moles. Since only 1 mole is needed per reaction it allows the reaction to go through 1.02 times.

LiOH molecular weight is 24. This gives 2.71 moles. Every reaction requires 2 moles so the reaction can go through 1.35 times.

Based on this, H₂SO₄ is the limiting reactant.

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Example Question #1 : Limiting Reagent

What determines the amount of product formed in an irreversible reaction?

Possible Answers:

The equilbrium state

The amount of limiting reactant present

The volume of the solution

The rate of the reverse reaction

Correct answer:

The amount of limiting reactant present

Explanation:

In irreversible reactions, the reaction proceeds in one direction only and these reactions usually go to completion. Since the forward reaction is the only one that occurs, the amount of product formed is only based on the reactants present, namely, the amount of limiting reactant.

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Example Question #4 : Limiting Reagent

Consider the following reaction:

$8Fe\hspace{1 mm}+S_8\hspace{1 mm}\rightarrow8FeS$

If we begin with 293 g Fe and 17.2 g $S_8$ , how many grams of $FeS$ will we create?

Possible Answers:

47.1 g

87.91 g

47.0 g

256.48 g

55.8 g

Correct answer:

47.1 g

Explanation:

First, calculate the theoretical yield as if we had 293 g $Fe$ and excess $S_8$ :

$293\hspace{1 mm}g\hspace{1 mm}Fe\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Fe}{55.8\hspace{1 mm}g\hspace{1 mm}Fe}\times\frac{8\hspace{1 mm}moles\hspace{1 mm}FeS}{8\hspace{1 mm}moles\hspace{1 mm}Fe}=5.25\hspace{1 mm}mol\hspace{1 mm}FeS$

Then calculate the theoretical yield as if we had 17.2 g $S_8$ :

$17.2\hspace{1 mm}g\hspace{1 mm}S_8\times\frac{1\hspace{1 mm}mole\hspace{1 mm}S_8}{256.48\hspace{1 mm}g\hspace{1 mm}S_8}\times\frac{8\hspace{1 mm}moles\hspace{1 mm}FeS}{1\hspace{1 mm}mole\hspace{1 mm}S_8}=5.36\times10^{-1}\hspace{1 mm}moles\hspace{1 mm} FeS$

The limiting reagent is $S_8$ as the theoretical yield calculation is lower. Now all there is to do is convert moles $FeS$ into grams:

$5.36\times10^{-1}\hspace{1 mm}moles\hspace{1 mm}FeS\times\frac{87.91\hspace{1 mm}g\hspace{1 mm}FeS}{1\hspace{1 mm}mole\hspace{1 mm}FeS}=47.1\hspace{1 mm}g\hspace{1 mm}FeS$

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Example Question #5 : Limiting Reagent

Consider the following reaction:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

If you have 36 grams of $Na$ and 53 mL $H_2O$ , will you have any remaining sodium and/or water? How much of each?

Possible Answers:

$0.106\hspace{1 mm}g\hspace{1 mm}Na$

$24.8\hspace{1 mm}mL\hspace{1 mm}H_2O$

$0.782\hspace{1 mm}g\hspace{1 mm}Na$

$28.2\hspace{1 mm}mL\hspace{1 mm}H_2O$

There will be no remaining water or sodium

Correct answer:

$24.8\hspace{1 mm}mL\hspace{1 mm}H_2O$

Explanation:

First, let us consider the 36 grams of solid sodium

$36\hspace{1 mm}g\hspace{1 mm}Na\times\frac{1\hspace{1 mm}mole\hspace{1 mm}Na}{22.99\hspace{1 mm}g\hspace{1 mm}Na}\times\frac{2\hspace{1 mm}moles\hspace{1 mm}H_2O}{2\hspace{1 mm}moles\hspace{1 mm}Na}\times\frac{18\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}mole\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mL\hspace{1 mm}H_2O}{1\hspace{1 mm}g\hspace{1 mm}H_2O}=28.2\hspace{1 mm}mL\hspace{1 mm}H_2O$

We now know that sodium is the limiting reagent as it uses only 28.2 mL of water and we have 53 mL of water. We will have no remaining sodium.

We will now calculate the remaining volume of water

$53\hspace{1 mm}mL-28.2\hspace{1 mm}mL=24.8\hspace{1 mm}mL\hspace{1 mm}H_2O$

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Example Question #1 : Limiting Reagent

Magnesium will combine with oxygen gas to form magnesium oxide according to the balanced equation below.

$2Mg + O_{2}\rightarrow2MgO$

60 grams of magnesium metal and 30 grams of oxygen gas are allowed to react. If the reaction runs to completion, which reactant will be depleted first?

Possible Answers:

Both will deplete at the same time.

More information is needed in order to answer the question.

Oxygen gas

Magnesium

Correct answer:

Oxygen gas

Explanation:

The reactant that will be depleted first is called the limiting reagent. We can determine the limiting reagent by calculating how much oxygen gas is necessary to use up all 60 grams of magnesium.

We start by converting the magnesium to moles.

$60g\ Mg*\frac{1mol\ Mg}{24.3g\ Mg}=2.47mol\ Mg$

We then compare the molar ratio of magnesium to oxygen gas. Since we need 1 mol of oxygen gas for every 2 moles of magnesium, we only need 1.24 moles of oxygen gas to fully react the magnesium.

$2.47mol\ Mg*\frac{1mol\ O_2}{2mol\ Mg}=1.23mol\ O_2$

Multiplying by the molar mass, we determine that we need 39.51 grams of oxygen gas. Remeber that the oxygen is diatomic in its gaseous state.

$1.23mol\ O_2*\frac{32g\ O_2}{1mol\ O_2}=39.51g\ O_2$

Since we only have 30 grams in the givevn question, we conclude that oxygen gas will be depleted first if the reaction runs to completion.

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Example Question #1 : Limiting Reagent

Consider the following balanced reaction.

$Cl_{2(g)} + 2KI_{(aq)} \rightarrow 2KCl_{(aq)} + I_{2}_{(s)}$

If the reaction starts with 8 moles of chlorine gas and 10 moles of potassium iodide, how many moles of the excess reagent will be left over after the reaction has run to completion?

Possible Answers:

$5\ mol$

$3\ mol$

$1\ mol$

$2\ mol$

Correct answer:

$3\ mol$

Explanation:

When reading the balanced reaction, you should notice that it requires 1 mol of chlorine gas and 2 moles of potassium iodide in order to react and create the products. This ratio is seen in the coefficients of the reagents.

Using this ratio, we can calculate the mole of chlorine needed to react with 10 moles of potassium iodide.

$10mol\ KI*\frac{1mol\ Cl_2}{2mol\ KI}=5mol\ Cl_2$

5 moles of chlorine gas will be used to react all of the potassium iodide. Since we started with 8 moles of chlorine gas, this means that there are 3 moles of chlorine gas left over after the reaction has run to completion.

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Example Question #1 : Limiting Reagent

Consider the following balanced reaction.

$Cl_{2(g)} + 2KI_{(aq)} \rightarrow 2KCl_{(aq)} + I_{2}_{(s)}$

If you start with 100 grams of chlorine gas and 500 grams of potassium iodide, how many grams of the excess reagent will remain after the reaction has run to completion?

Possible Answers:

$\small 32 g$

$\small 300 g$

$\small 266 g$

$\small 64 g$

Correct answer:

$\small 32 g$

Explanation:

Since we are given masses in this problem, it is a little trickier to compare the amounts of both reagents to one another. We can use stoichiometry to determine how much of one reagent is needed in order to use up all of the other reagent. We can determine how much potassium iodide is necessary to use up all of the chlorine gas by using the following series of calculations.

Convert both compounds to moles.

$100g\ Cl_2*\frac{1mol}{71g}=1.41mol\ Cl_2$

$500g\ KI*\frac{1mol}{166g}=3.01mol\ KI$

Then, use the ratio in the chemical reaction to determine the moles of potassium iodide needed to react all of the chlorine.

$1.41mol\ Cl_2*\frac{2mol\ KI}{1mol\ Cl_2}=2.82mol\ KI$

Only 2.82 moles of poatssium iodide are needed. Since we started with 3.01 moles of potassium iodide, 0.19 moles will be left over. Now, we need to convert this remainsing amount to grams.

$0.19mol\ KI*\frac{166g}{1mol}=32g\ KI$

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AP Chemistry : Stoichiometry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #11 : Balancing Equations

Transesterification is an industrially important process for the production of biodiesel fuel (that most diesel engines can run cleanly on) and glycerin from triglycerides (usually people use vegetable oil as their feedstock).

Example Question #431 : Ap Chemistry

Example Question #1 : Limiting Reagent

Example Question #1 : Limiting Reagent

Example Question #1 : Limiting Reagent

Example Question #4 : Limiting Reagent

Example Question #5 : Limiting Reagent

Example Question #1 : Limiting Reagent

Example Question #1 : Limiting Reagent

Example Question #1 : Limiting Reagent

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