Chemical equations Stoichiometry problems

Calculations like this are called equation stoichiometry problems, or just stoichiometry problems. Stoichiometry, from the Greek words for measure and element, refers to the quantitative relationships between substances. Calculations such as those in Chapter 9, in which we convert between an amount of compound and an amount of element in the compound, are one kind of stoichiometry problem, but the term is rarely used in that context. Instead, it is usually reserved for calculations such as the one above, which deal with the conversion of the amount of one substance in a chemical reaction into the amount of a different substance in the reaction. 

To see how molarity can be used in equation stoichiometry problems, let s take a look at the thought process for calculating the number of milliliters of 1.00 M AgN03 necessary to precipitate the phosphate from 25.00 mL of0.500 M Na3P04. The problem asks us to convert from amount of one substance in a chemical reaction to amount of another substance in the reaction, so we know it is an equation stoichiometry problem. The core of our setup will be the conversion factor for changing moles of sodium phosphate to moles of silver nitrate. To construct it, we need to know the molar ratio of AgN03 to Na3P04, which comes from the balanced equation for the reaction. 

Because we are asked to convert from the amount of one substance in a chemical reaction to amount of another substance in the reaction, we recognize the problem as an equation stoichiometry problem. Our general steps are... 

Because we are converting from units of one substance to units of another substance, both involved in a chemical reaaion, we recognize this problem as an equation stoichiometry problem. We can therefore use unit analysis for our calculation. 

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 ... 

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. 

The first step in any stoichiometry problem is to write the balanced chemical equation ... 

In working stoichiometry problems you will need the balanced chemical equation. In addition, if the problem involves a quantity other than moles, you will need to convert to moles. 

In stoichiometry problems, be sure to use the balanced chemical equation. 

The sequence of conversions in Figure 18.20 is used to calculate the mass or volume of product produced by passing a known current through a cell for a fixed period of time. The key is to think of the electrons as a "reactant" in a balanced chemical equation and then to proceed as with any other stoichiometry problem. Worked Example 18.10 illustrates the calculations. Alternatively, we can calculate the current (or time) required to produce a given amount of product by working through the sequence in Figure 18.20 in the reverse direction, as shown in Worked Example 18.11. 

Occasionally, not all equilibrium concentrations are known. When this occurs you must use equilibrium concepts and stoichiometry concepts to determine K. What you are trying to do in these problems is determine the amounts of materials at equilibrium. In Chapter 12, you learned that the balanced chemical equation shows you the relative amounts of reactants and products during the chemical reaction. For a reaction at equilibrium, the logic is the same. The mole ratios still apply. There is one major difference, however, between the stoichiometry... 

Recall that stoichiometry involves calculating the amounts of reactants and products in chemical reactions. If you know the atoms or ions in a formula or a reaction, you can use stoichiometry to determine the amounts of these atoms or ions that react. Solving stoichiometry problems in solution chemistry involves the same strategies you learned in Unit 2. Calculations involving solutions sometimes require a few additional steps, however. For example, if a precipitate forms, the net ionic equation may be easier to use than the chemical equation. Also, some problems may require you to calculate the amount of a reactant, given the volume and concentration of the solution. 

EXAMPLE 10.1. How many moles of AI2O3 can be prepared from the reaction of 0.450 mol of O2 pins sufficient Al Ans. The first step in any stoichiometry problem is to write the balanced chemical equation ... 

Substances are usually measured by mass or volume. As a result, before using the mole ratio you will often need to convert between the units for mass and volume and the unit mol Yet each stoichiometry problem has the step in which moles of one substance are converted into moles of a second substance using the mole ratio from the balanced chemical equation. Follow the steps in Skills Toolkit 2 to understand the process of solving stoichiometry problems. 

Stoichiometry problems can be solved using three basic steps. First, change what you are given into moles. Second, use a mole ratio based on a balanced chemical equation. Third, change to the units needed for the answer. 

Why is a balanced chemical equation needed to solve stoichiometry problems ... 

Stoichiometry concerns calculations based on balanced chemical equations, a topic that was presented in Chapter 8. Remember that the coefficients in the balanced equations indicate the number of moles of each reactant and product. Because many reactions take place in solution, and because the molarity of solutions relates to moles of solute and volumes, it is possible to extend stoichiometric calculations to reactions involving solutions of reactants and products. The calculations involving balanced equations are the same as those done in Chapter 8, but with the additional need to do some molarity calculations. Let s get our feet wet by working a couple of problems involving solutions in chemical reactions. 

Stoichiometry is the subject that tells the quantity of one substance that reacts with some quantity of anything else in a chemical reaction. The coefficients of the balanced chemical equation give the mole ratios of every substance in the reaction to every other substance. Therefore it is imperative to write a balanced chemical equation for every problem involving a chemical reaction. The equation... 

Conversions of moles of one substance to moles of any other in the balanced chemical equation is straightforward just remember that it is moles not mass that is related to the coefficients in the equation, ffowever, stoichiometry problems often give students more trouble than they should because the problems are often asked in terms of masses or other quantities that can be related to moles of reactant or product. These problems are multistep problems, but should not present too much difficulty because each individual step is straightforward. 

Note that the concentrations of B and D are squared, as required by the equilibrium constant expression. Note also that we got those concentrations from the stoichiometry of the problem, and we did not double the B or D concentration. The D concentration happens to be twice the C concentration, but it is the D concentration that we square. Please also note that the "Change due to reaction" row in the final table is the only row that is in the ratio of the balanced chemical equation. ... 

When solving stoichiometry problems for solution reactions, what type of chemical equation is most convenient to use ... 

We need to work backwards from the final goal to decide where to start. For example, in a stoichiometry problem we always start with the chemical reaction. Then we ask a series of questions as we proceed, such as, "What are the reactants and products " "What is the balanced equation " "What are the amounts of the reactants " and so on. Our understanding of the fundamental principles of chemistry will enable us to answer each of these simple questions and will eventually lead us to the final solution. We might summarize this process as "How do we get there "... 

Two factors enter into this problem. The first is the stoichiometry of the reaction transforming A to B. Chemical equations as normally written express the... 

What happens when the stoichiometric relationships are not one-to-one For example, consider the reaction 2 HI(g) H2(g) + Iilg)- We can measure either the rate of disappearance of HI or the rate of appearance of either H2 or I2. Because 2 mol of HI disappear for each mole of H2 or I2 that forms, the rate of disappearance of HI is twice the rate of appearance of either H2 or I2. How do we decide which number to use for the rate of the reaction Depending on whether we monitor HI, I2, or H2, the rates can differ by a factor of two. To fix this problem, we need to take into account the reaction stoichiometry. To arrive at a number for the reaction rate that does not depend on which component we measured, we must divide the rate of disappearance of HI by 2 (its coefficient in the balanced chemical equation) ... 

Stoichiometry is a term used to describe quantitative relationships in chemistry. Any chemistry question that asks How much of a particular substance will be consumed or formed in a given chemical reaction is a stoichiometry problem. At the heart of every such stoichiometry problem, you ll always find a balanced chemical equation. 

The heart of any stoichiometry problem is the balanced chemical equation that provides the mole ratio we need. We must convert between masses and the number of moles in order to use this ratio, first for the reactant, 5, and at the end of the problem for the product, P4S3. Molar mass is used to provide the needed conversion factors. The phrase, excess of phosphorus, tells us that we have more than enough P4 to consume 153 g of 8 completely. 

A titration problem is an applied stoichiometry problem, so we will need a balanced chemical equation. We know the molarity and volume for the NaOH solution, so we can find the number of moles reacting. The mole ratio from the balanced equation lets us calculate moles of H2SO4 firom moles of NaOH. Because we know the volume of the original H2SO4 solution, we can find its molarity. 

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