Balanced equations limiting reactant determination

Write a balanced chemical equation for the reaction. Find the amount (in mol) of each reactant, using its volume and concentration. Identify the limiting reactant. Determine the amount (in mol) of mercury(II) sulfide that forms. Calculate the mass of mercury(II) sulfide that precipitates. 

How can you determine which reactant in a chemical reaction is limited First, find the number of moles of each reactant by multiplying the given mass of each reactant by the inverse of the molar mass. Next, determine whether the reactants are available in the mole ratio specified in the balanced equation. A reactant that is available in an amount smaller than that required by the mole ratio is a limiting reactant. 

C04-0041. Several examples of chemical reasoning are introduced in this chapter. Write out the reasoning steps that you will follow in (a) balancing a chemical equation (b) identifying the limiting reactant (c) determining whether a precipitate forms and (d) computing a reaction yield. 

When two substances react, they react in exact amounts. You can determine what amounts of the two reactants are needed to react completely with each other by means of mole ratios based on the balanced chemical equation for the reaction. In the laboratory, precise amounts of the reactants are rarely used in a reaction. Usually, there is an excess of one of the reactants. As soon as the other reactant is used up, the reaction stops. The reactant that is used up is called the limiting reactant. Based on the quantities of each reactant and the balanced chemical equation, you can predict which substance in a reaction is the limiting reactant. 

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. 

In this chapter, you learned how to balance simple chemical equations by inspection. Then you examined the mass/mole/particle relationships. A mole has 6.022 x 1023 particles (Avogadro s number) and the mass of a substance expressed in grams. We can interpret the coefficients in the balanced chemical equation as a mole relationship as well as a particle one. Using these relationships, we can determine how much reactant is needed and how much product can be formed—the stoichiometry of the reaction. The limiting reactant is the one that is consumed completely it determines the amount of product formed. The percent yield gives an indication of the efficiency of the reaction. Mass data allows us to determine the percentage of each element in a compound and the empirical and molecular formulas. 

First of all, we have to determine which the limiting reactant is. The way to determine which reactant is limiting is to divide the moles of each reactant by the coefficient from the balanced equation associated with that reactant. The smallest number that comes out indicates which reactant is the limiting one. This reactant limits how much of every other species made or needed for the reaction. For A, this calculation gives 0.25 and for B, 0.5. Thus, A is the limiting reactant and the calculation of d, sR, and x should based on it. 

It s fairly easy to conceptualize the idea of limiting reactants when you are given moles of the reactants. When you are given grams, it is not always so easy to see. When you have to solve limiting reactant problems, it is always necessary to determine the number of moles of each substance and compare that to the required ratios from the balanced chemical equation. Let s use the same reaction, but use masses instead of 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 ... 

The following balanced chemical equation shows the reaction of aluminum with copper(II) chloride. If 0.25 g of aluminum reacts with 0.51 g of copper(II) chloride, determine the limiting reactant. 

When a reaction of two or more substances has proceeded until one of the reactants has been used up, the reaction stops. To determine which reactant is limiting, divide the number of moles of each by its coefficient in the balanced equation. The lowest quotient is that of the limiting quantity. Do not use these quotients for any further calculations. 

A related, but simpler, way to determine which reactant is limiting is to compare the mole ratio of the substances required by the balanced equation with the mole ratio of reactants actually present. For example, in this case the mole ratio of H2 to N2 required by the balanced equation is... 

By comparing the mole ratio of reactants required by the balanced equation with the mole ratio of reactants actually present, determine which reactant is limiting. 

In Chapter 3 we covered the principles of chemical stoichiometry the procedures for calculating quantities of reactants and products involved in a chemical reaction. Recall that in performing these calculations, we first convert all quantities to moles and then use the coefficients of the balanced equation to assemble the appropriate molar ratios. In cases in which reactants are mixed, we must determine which reactant is limiting, since the reactant that is consumed first will limit the amounts of products formed. These same principles apply to reactions that take place in solutions. However, there are two points about solution reactions that need special emphasis. The first is that it is sometimes difficult to tell immediately which reaction will occur when two solutions are mixed. Usually we must think about the various possibilities and then decide what will happen. The first step in this process always should be to write down the species that are actually present in the solution, as we did in Section 4.5. 

The calculations you did in Section 12.2 were based on having the reactants present in the ratio described by the balanced chemical equation. How can you calculate the amount of product formed when one reactant limits the amount of product and the other is in excess The first thing you must do is determine which reactant is the limiting reactant. 

Given a set of initial masses of reactants and a balanced chemical equation, determine the limiting reactant and calculate the masses of reactants and products after the reaction has gone to completion (Section 2.6, Problems 47 and 48). 

To solve many of the problems in this chapter, you will need to apply the limiting reactant concept (Section 3-3). In Example 11-1, we confirm that the two reactants are initially present in the mole ratio required by the balanced chemical equation they both react completely, so there is no excess of either one. In Example 11-2, we need to determine which reactant limits the reaction. Before you proceed, be sure you understand how the ideas of Section 3-3 are used in these examples. 

We must determine how much reaction occurred—that is, how many moles of reactants were consumed. We first multiply the volume, in liters, of each solution by its concentration in mol/L (molarity) to determine the number of moles of each reactant mixed. Then we identify the limiting reactant. We scale the amount of heat released in the experiment to correspond to the number of moles of that reactant shown in the balanced equation. 

For our reaction between carbon and silicon dioxide, we can determine which reactant is the limiting reactant by first calculating the maximum amount of silicon that can be formed from the given amount of each reactant. The reactant that forms the least product will run out first and limit the amount of product that can form. The coefficients in the balanced equation provide us with conversion factors to convert from moles of reactants to moles of products. 

To determine which is the limiting reactant, choose one of the reactants and determine how many moles of the other it will consume. Let s choose NaOH and ask the question How many moles of HCl will be consumed by 0.35 mole of NaOH Even without doing a mole-to-mole conversion, the 1-to-l mole ratio in the balanced equation indicates that... 

The equation is balanced and starting amounts of two reactants are given. Determine the limiting reactant, then calculate the mass of NiS produced based on the limiting reactant. The necessary molar masses are calculated NiCl2 = 129.6 g (NH4)2S = 68.15 g NiS = 90.76 g. 

Plan We first write the balanced equation. Because the amounts of two reactants are given, we know this is a limiting-reactant problem. To determine which reactant is limiting, we calculate the mass of N2 formed from each reactant assuming an excess of the other. We convert the grams of each reactant to moles and use the appropriate molar ratio to find the moles of N2 each forms. Whichever yields less N2 is the limiting reactant. Then, we convert this lower number of moles of N2 to mass. The roadmap shows the steps. Solution Writing the balanced equation ... 

Plan This is a limiting-reactant problem because the amounts of two reactants are given. After balancing the equation, we must determine the limiting reactant. The molarity (0.010 M) and volume (0.050 L) of the mercury(II) nitrate solution tell us the moles of one reactant, and the molarity (0.10 M) and volume (0.020 L) of the sodium sulfide solution tell us the moles of the other. Then, as in Sample Problem 3.10, we use the molar ratio to find the moles of HgS that form from each reactant, assuming the other reactant is present in excess. The limiting reactant is the one that forms fewer moles of HgS, which we convert to mass using the HgS molar mass. The roadmap shows the process. 

Step 3 Now we determine which reactant is limiting. Using the mole ratio between CO and H2 from the balanced equation, we have... 

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