Limiting Reagent - AP Chemistry

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Question

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

Answer

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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Question

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?

Answer

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 O2, 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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

If you start the reaction with 12g of hydrogen gas and 64g of nitrogen gas, which will be the limiting reagent?

Answer

When finding the limiting reactant, it is a good idea to take one of the reactants and solve for how much of the other will be needed to fully react it. For example, let us calculate how much nitrogen gas we would need in order to use up all 12g of hydrogen gas. Using the equation to find moles, we find that we have 6mol of hydrogen gas.

$12gH_2\frac{1molH_2}{2gH_2}=6molH_2$

Using the molar ratio of 1:3 from the chemical equation, we conclude that we only need 2mol of nitrogen gas to use up all of the hydrogen gas.

$6molH_2\frac{1molN_2}{3molH_2}=2molN_2$

Multiplying 2mol of nitrogen gas by its molar mass (28 grams per mol) gives us 56g of nitrogen gas needed to react all of the hydrogen.

$2molN_228\frac{g}{mol}N_2=56gN_2$

Since we started with 64g of nitrogen gas, we know that we have more than enough and that hydrogen will be used up first.* Hydrogen gas is the limiting reactant.

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Question

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?

Answer

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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Question

How is a limiting reagent problem recognized?

Answer

When two or more reactants of a chemical equation's quantities are given, the first step is to determine the limiting reagent of the reaction. The reaction will only proceed and produce as much product as the limiting reagent allows. For example if there are 5 moles of oxygen gas and excess hydrogen gas, only 10 moles of water can be produced regardless of how much hydrogen gas is present.

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Question

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?

Answer

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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Question

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?

Answer

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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Question

Consider the reaction of potassium carbonate with calcium nitrate to form potassium nitrate and calcium carbonate:

$K_{2}CO_{3} (aq) + Ca(NO_{3})_{2} (aq) \rightarrow CaCO_{3} (s) + 2KNO_{3} (aq)$

Suppose 50ml of a 0.250M potassium carbonate solution was mixed with 100ml of a 0.175M calcium nitrate solution. What is the maximum amount of calcium carbonate that could be obtained?

Answer

First, we must determine how many moles of each reactant begin the reaction by multiplying the molarity by the volume. Don't forget to convert volume to liters!

$(0.250M)(0.050L)=0.0125mol\ K_2CO_3$

$(0.175M)(0.100L)=0.0175mol\ Ca(NO_3)_2$

Next, use the reaction coefficients (i.e. the stoichiometry) to determine how many moles of calcium carbonate could be formed from each of the reactants. In this case, there is a 1:1 molar ratio between both reactants and calcium carbonate.

$0.0125mol\ K_2CO_3\frac{1mol\ CaCO_3}{1mol\ K_2CO_3}=0.0125mol\ CaCO_3$

$0.0175mol\ Ca(NO_3)_2\frac{1mol\ CaCO_3}{1mol\ Ca(NO_3)_2}=0.0175mol\ CaCO_3$

Thus, 0.0125 moles of potassium carbonate could form 0.0125 moles of calcium carbonate, while 0.0175 moles of calcium nitrate could form 0.0175 moles of calcium carbonate. The maximum amount of product is going to be determined by the limiting reactant, i.e. the reactant that provides the least amount of product. In this case, the limiting reactant is potassium carbonate, and the maximum yield of calcium carbonate is 0.0125mol.

For the final step convert this value to grams:

$0.0125mol\ CaCO_3*\frac{100.01g\ CaCO_3}{1mol\ CaCO_3}=1.25g\ CaCO_3$

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Question

The following reaction is used to obtain small amounts of chlorine gas in the laboratory:

$MnO_{2(s)}: +: 4HCl_{(g)}: \rightarrow : MnCl_{2(s)}: +: Cl_{2(g:)}+: 2H_{2}O_{(l)}$

If of $MnO_{2}$ are allowed to react with of , the limiting reactant will be:

Answer

First of all, a product cannot be the limiting reactant. This problem can be solved in many different ways. Keep in mind that to solve a stoichiometry problem a good practice is to convert mass to mol. Let's find if the amount of we have is enough to react with of $MnO_{2}$ :

$28.0: : g: : of: MnO_{2}\times \frac{1: mol: MnO_{2}}{86.94: g : MnO_{2}}\times \frac{4: mol: HCL}{1: mol: MnO_{2}}\times \frac{36.46 : g: HCL}{1: mol: HCL}=$

We have only of and are needed. Then is the limiting reactant.

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Question

When grilling out, many people utilize the combustion of propane to provide the heat energy needed to cook their food. The chemical equation for this reaction is shown below:

$C_3H_8_{(g)}+5O_2_{(g)}\rightarrow 3CO_2_{(g)}+4H_2O_{(g)}$

You notice that your gas grill is producing a large amount of soot, which is negatively impacting the taste of your food. Please select the best explanation of this phenomenon, and it's corresponding solution.

Answer

Soot is a mixture incompletely combusted hydrocarbons. Therefore the production of soot occurs most when the hydrocarbon is in excess and the oxidizer (in this case oxygen) is the limiting reactant.

Incorrect answers and explanations:

The limiting reactant is propane, therefore the best way to reduce the soot is to increase the input of propane.

This is incorrect because propane is in excess when soot production is maximized. Therefore adding more propane will only result in more soot.

The limiting reactant is oxygen; therefore the best way to reduce the soot is to add an oxygen source such as , which releases oxygen by thermal decomposition.

The limiting reactant is indeed oxygen and does undergo thermal decomposition to produce oxygen. However it is also one of the primary ingredients for gunpowder (the other two ingredients have a high probability of being present in a device that processes tissue from living organisms), therefore the safety factor of this answer leads it to be incorrect. Not to mention that this method would not be the simplest solution.

Oxygen is in excess; therefore the best way to reduce the soot is to close the grill hood and close any vent holes preventing atmospheric oxygen from entering the system.

Oxygen is the limiting reactant; therefore cutting off the supply of oxygen would only create more soot or depending upon the degree to which the oxygen supply is cut off, this could stop the reaction altogether.

The soot production is due to a dirty grill and therefore your grill must be thoughly cleaned.

A dirty grill may have soot in it, but unless the dirtiness is blocking off the supply of oxygen, the soot currently being produced is not caused by the dirt. In any case the root cause of any soot production is that oxygen is limiting and this answer does not address that fact.

Correct Answer and explanation:

The limiting reactant is oxygen; therefore the best way to reduce the soot is to decrease the input of propane, which is in excess.

The limiting reactant when soot is produced is oxygen. In most propane grills, the oxygen supply is maxed out at the level of atmospheric oxygen. Therefore the only reactant under your direct control is the propane. By reducing the quantity of propane, you can make it no longer in excess.

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Question

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

Answer

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.

Compare your answer with the correct one above

Question

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?

Answer

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 O2, 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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Question

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?

Answer

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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Question

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?

Answer

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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Question

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?

Answer

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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Question

The formation of ammonia is given by the following equation:

$3H_2+N_2\rightarrow 2NH_3$

If you start the reaction with 12g of hydrogen gas and 64g of nitrogen gas, which will be the limiting reagent?

Answer

When finding the limiting reactant, it is a good idea to take one of the reactants and solve for how much of the other will be needed to fully react it. For example, let us calculate how much nitrogen gas we would need in order to use up all 12g of hydrogen gas. Using the equation to find moles, we find that we have 6mol of hydrogen gas.

$12gH_2\frac{1molH_2}{2gH_2}=6molH_2$

Using the molar ratio of 1:3 from the chemical equation, we conclude that we only need 2mol of nitrogen gas to use up all of the hydrogen gas.

$6molH_2\frac{1molN_2}{3molH_2}=2molN_2$

Multiplying 2mol of nitrogen gas by its molar mass (28 grams per mol) gives us 56g of nitrogen gas needed to react all of the hydrogen.

$2molN_228\frac{g}{mol}N_2=56gN_2$

Since we started with 64g of nitrogen gas, we know that we have more than enough and that hydrogen will be used up first.* Hydrogen gas is the limiting reactant.

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Question

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?

Answer

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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