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Example Questions
Example Question #431 : Ap Chemistry
Consider the following unbalanced reaction:
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)?
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.
Because there are moles of on the right side, we'll need to have moles of . Next, let's go ahead and balance nitrogen on both sides.
Since we added a coefficient of to the in the first step, there are now moles of on the left. Thus, we'll need to add a coefficient of to the . Next, if we check the amount of , , and on both sides, we see that they are equal. Thus, we can recognize that and have coefficients of , and do not need to be altered.
Our final balanced equation looks like this.
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?
Ca because there will need to be 2 moles if Ca for every mole of O2 as defined by the balanced equation.
O2 because it is a diatomic molecule
There will not be a limiting reactant
O2 because there is less of it
Ca because there are 1.5 moles of Ca and only 1 mole of O2
Ca because there will need to be 2 moles if Ca for every mole of O2 as defined by the balanced equation.
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.
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?
H2SO4
Li2SO4
Both H2SO4 and LiOH limit the reaction equally
H2O
LiOH
H2SO4
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.
Example Question #3 : Limiting Reagent
What determines the amount of product formed in an irreversible reaction?
The amount of limiting reactant present
The volume of the solution
The rate of the reverse reaction
The equilbrium state
The amount of limiting reactant present
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.
Example Question #1 : Limiting Reagent
Consider the following reaction:
If we begin with 293 g Fe and 17.2 g , how many grams of will we create?
47.1 g
256.48 g
47.0 g
87.91 g
55.8 g
47.1 g
First, calculate the theoretical yield as if we had 293 g and excess :
Then calculate the theoretical yield as if we had 17.2 g :
The limiting reagent is as the theoretical yield calculation is lower. Now all there is to do is convert moles into grams:
Example Question #2 : Limiting Reagent
Consider the following reaction:
If you have 36 grams of and 53 mL , will you have any remaining sodium and/or water? How much of each?
There will be no remaining water or sodium
First, let us consider the 36 grams of solid sodium
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
Example Question #3 : Limiting Reagent
Magnesium will combine with oxygen gas to form magnesium oxide according to the balanced equation below.
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?
Oxygen gas
More information is needed in order to answer the question.
Magnesium
Both will deplete at the same time.
Oxygen gas
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.
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.
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.
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.
Example Question #4 : Limiting Reagent
Consider the following balanced reaction.
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?
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.
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.
Example Question #5 : Limiting Reagent
Consider the following balanced reaction.
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?
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.
Then, use the ratio in the chemical reaction to determine the moles of potassium iodide needed to react all of the chlorine.
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.
Example Question #1 : Limiting Reagent
The formation of ammonia is given by the following equation:
If you start the reaction with 12g of hydrogen gas and 64g of nitrogen gas, which will be the limiting reagent?
We must know the equilibrium constant in order to solve
Nitrogen
Both will be used up at the same time
We must know the system pressure in order to solve
Hydrogen
Hydrogen
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.
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.
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.
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.