Stoichiometry problems limiting reactants

We have data for the amounts of both starting materials, so this is a limiting reactant problem. Given the chemical equation, the first step in a limiting reactant problem is to determine the number of moles of each starting material present at the beginning of the reaction. Next compute ratios of moles to coefficients to identify the limiting reactant. After that, a table of amounts summarizes the stoichiometry. 

Stoichiometry problems (including limiting-reactant problems) involving solutions can be worked in the same fashion as before, except that the volume and molarity of the solution must first be converted to moles. 

The correct answer is (A). When you see two masses in a stoichiometry problem, you should be alerted that you are dealing with a limiting reactant problem. This problem will have two stages—the first is to determine the limiting reactant, and the second to determine the mass of the hydrogen gas. Before we do anything, we need to see the balanced equation for the reaction ... 

O How is a limiting reactant problem different from other stoichiometry problems (What is your clue that it is a limiting reactant problem )... 

Chapters 3 and 4 include more extensive and consistent use of stoichiometry reaction tables in limiting-reactant problems. 

Whenever you are confronted with a stoichiometry problem you should always determine if you are going to have to solve a limiting reactant problem like this one, or a problem like Example 3.13 that involves a single reactant and one reactant in excess. A good rule of thumb is that when two or more reactant quantities are specified, you should approach the problem as was done here. 

A table of amounts is a convenient way to organize the data and summarize the calculations of a stoichiometry problem. Such a table helps to identify the limiting reactant, shows how much product will form during the reaction, and indicates how much of the excess reactant will be left over. A table of amounts has the balanced chemical equation at the top. The table has one column for each substance involved in the reaction and three rows listing amounts. The first row lists the starting amounts for all the substances. The second row shows the changes that occur during the reaction, and the last row lists the amounts present at the end of the reaction. Here is a table of amounts for the ammonia example ... 

The problem asks for a yield, so we identify this as a yield problem. In addition, we recognize this as a limiting reactant situation because we are given the masses of both starting materials. First, identify the limiting reactant by working with moles and stoichiometric coefficients then carry out standard stoichiometry calculations to determine the theoretical amount that could form. A table of amounts helps organize these calculations. Calculate the percent yield from the theoretical amount and the actual amount formed. 

The quantitative treatment of a reaction equilibrium usually involves one of two things. Either the equilibrium constant must be computed from a knowledge of concentrations, or equilibrium concentrations must be determined from a knowledge of initial conditions and Kgq. In this section, we describe the basic reasoning and techniques needed to solve equilibrium problems. Stoichiometry plays a major role in equilibrium calculations, so you may want to review the techniques described in Chapter 4, particularly Section 4- on limiting reactants. 

Again, we use the standard approach to an equilibrium calculation. In this case the reaction is a precipitation, for which the equilibrium constant is quite large. Thus, taking the reaction to completion by applying limiting reactant stoichiometry is likely to be the appropriate approach to solving the problem. 

Complex stoichiometry problems should be worked slowly and carefully, one step at a time. When solving a problem that deals with limiting reactants, the idea is to find how many moles of all reactants are actually present and then compare the mole ratios of those actual amounts to the mole ratios required by the balanced equation. That comparison will identify the reactant there is too much of (the excess reactant) and the reactant there is too little of (the limiting reactant). 

Assume that your friend has missed several chemistry classes and that she has asked you to help her prepare for a stoichiometry test. Unfortunately, because of other commitments, you do not have time to meet face to face. You agree to email your friend a set of point-form instructions on how to solve stoichiometry problems, including those that involve a limiting reactant. She also needs to understand the concept of percentage yield. Write the text of this email. Assume that your friend has a good understanding of the mole concept. 

If, however, 2.50 X 103 kilograms of methane is mixed with 3.00 X 103 kilograms of water, the methane will be consumed before the water runs out. The water will be in excess. In this case the quantity of products formed will be determined by the quantity of methane present. Once the methane is consumed, no more products can be formed, even though some water still remains. In this situation, because the amount of methane limits the amount of products that can be formed, it is called the limiting reactant, or limiting reagent. In any stoichiometry problem it is essential to determine which reactant is the limiting one to calculate correctly the amounts of products that will be formed. 

In the problems we have worked thus far, the presence of an excess of one reactant was stated or implied. The calculations were based on the substance that was used up first, called the limiting reactant. Before we study the concept of the limiting reactant in stoichiometry, let s develop the basic idea by considering a simple but analogous nonchemical example. 

For this calculation, we follow a procedure that is very similar to the procedure we would follow to calculate the moles of silicon that can be made from the reaction of 1000 moles of carbon and 550 moles of Si02. We calculate the amount of silicon that can be made from 16.491 g C and also from 32.654 g Si02. Whichever forms the least product is the limiting reactant and therefore determines the maximum amount of product that can form. These two calculations are equation stoichiometry problems, so we will use the procedure described in Sample Study Sheet 10.1. 

If NaOH and HCl(aq) are combined in anything other than a 1 to 1 mole ratio, one reactant will be in excess. The other will be the limiting reactant. Whenever starting amounts of two reactants are given in a stoichiometry problem, before you can calculate amounts of product you will first have to learn which reactant is the limiting reactant. It will govern how much product forms. 

Steps for Solving Stoichiometry Problems Involving Limiting Reactants... 

As for any reaction stoichiometry problem, we should start with a balanced equation. One way to proceed from there is to calculate the amount of one reactant that would combine with the given amount of the second reactant. Comparing that with the amount actually available will reveal the limiting reactant. 

This problem combines two types of calculations that we ve seen in earlier examples. The first step is identifying the limiting reactant, and the second is the actual stoichiometry calculation. Note that here isobutene was the limiting reactant even though the available mass of isobutene was greater than that of methanol. This is a reminder that reaction stoichiometry calculations should always be handled in terms of moles, not masses. 

We are asked for the expected amount of a product, so this is a reaction stoichiometry problem. Because NH3 is said to be in excess, we know that NaClO will be the limiting reactant. So we will use the given volume and concentration to find the number of moles of NaClO reacting. Then use the mole ratio Irom the balanced equation to find the number of moles of N2H4 that can be formed. Finally, we can use that number of moles and the given final volume to obtain the molarity. 

All stoichiometry problems can be approached with this general pattern. In some cases, however, additional calculations may be needed. For example, if we are given (or able to measure) known amounts of two or more reactants, we must determine which of them will be completely consumed (the limiting reactant.) Once again, the mole ratios in the balanced equation hold the key. Another type of calculation that can be considered in a stoichiometry problem is determining the percentage yield of a reaction. In this case, the amount of product determined in the problem... 

First, what if carbon is the limiting reactant How many grams of COj will be produced In other words, if all the carbon is used up in the reaction, how many grams of CO2 will be made This is a fundamental three-step stoichiometry problem ... 

As with other problems with stoichiometry, it is the less abundant reactant that limits the product. Accordingly, we define the extent of reaction p to be the fraction of A groups that have reacted at any point. Since A and B groups... 

This is a critical chapter in your study of chemistry. Our goal is to help you master the mole concept. You will learn about balancing equations and the mole/mass relationships (stoichiometry) inherent in these balanced equations. You will learn, given amounts of reactants, how to determine which one limits the amount of product formed. You will also learn how to determine the empirical and molecular formulas of compounds. All of these will depend on the mole concept. Make sure that you can use your calculator correctly. If you are unsure about setting up problems, refer back to Chapter 1 of this book and go through Section 1-4, on using the Unit Conversion Method. Review how to find atomic masses on the periodic table. Practice, Practice, Practice. 

The amount of product is calculated by the same method used earlier for mole-to-gram stoichiometry problems. Start with the moles of the limiting factor because a limiting factor is defined as the reactant that limits or determines the amount of product that can be made. 

The most important characteristic of this problem is that the Hougen-Watson kinetic model contains molar densities of more than one reactive species. A similar problem arises if 5 mPappl Hw = 2CaCb because it is necessary to relate the molar densities of reactants A and B via stoichiometry and the mass balance with diffusion and chemical reaction. When adsorption terms appear in the denominator of the rate law, one must use stoichiometry and the mass balance to relate molar densities of reactants and products to the molar density of key reactant A. The actual form of the Hougen-Watson model depends on details of the Langmuir-Hinshelwood-type mechanism and the rate-limiting step. For example, consider the following mechanism ... 

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