Stoichiometry solutions

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. 

Problem Is How many milliliters of 0.500 M sulfuric acid, H2SO4(aq), are needed to just completely react with 12.0 g of sodium hydroxide, NaOH The molar mass of NaOH is 40.0 g, and the balanced equation for the reaction is 

After the number of moles of NaOH is calculated, the balanced equation will be used to determine the required number of moles of acid. The pathway to the answer is 

Number of mole of NaOH = 12.0 x (1 mole NaOH/40.0 j ) = 0.300 mole NaOH. 

The balanced equation shows that 2 moles of NaOH react with 1 mole of sulfuric acid. We are seeking the mole of acid that will just consume 0.300 mole NaOH. The coefficient ratio of sought (1 mole H2S04) over known (2 mole NaOH) equals (1/2). 

If 25 g of salt are dissolved in 251 g of water, what is the mass of the resulting solution  

As we discussed in Chapter 7, many chemical reactions take place in aqueous solutions. Precipitation reactions, neutralization reactions, and gas evolution reactions, for example, all occur in aqueous solutions. Chapter 8 describes how we use the coefficients in chemical equations as conversion factors between moles of reactants and moles of products in stoichiometric calculations. These conversion factors are often used to determine, for example, the amount of product obtained in a chemical reaction based on a given amount of reactant or the amount of one reactant needed to completely react with a given amount of another reactant. The general solution map for these kinds of calculations is  

In reactions involving aqueous reactant and products, it is often convenient to specify tiie amoxmt of reactants or products in terms of their volume and concentra-hon. We can use the volume and concentrahon to calculate Ihe number of moles of reactants or products, and then use tire stoichiometric coefficients to convert to other quanhties in the reachon. The general soluhon map for these kinds of calculations is  

How much 0.125 M NaOH solution do we need to completely neutralize 0.225 L of 0.175 M H2SO4 solution Begin by sorting the information in the problem statement. 

We strategize by drawing a solution map, which is similar to those for other stoichiometric problems. We first use the volume and molarity of H2SO4 solution to get mol H2SO4. Then we use the stoichiometric coefficients from the equation to convert mol H2SO4 to mol NaOH. Finally, we use the molarity of NaOH to get to L NaOH solution. 

In the case of a soluble ionic compound with other than a 1 1 combination of constituent ions, we must use the subscripts in the chemical formula to determine the concentration of each ion in solution. Sodium sulfate (Na2S04) dissociates, for example, to give twice as many sodium ions as sulfate ions. 

Therefore, a solution that is 0.35 M in Na2S04 is actually 0.70 M in Na and 0.35 M in SOj . Frequently, molar concentrations of dissolved species are expressed using square brackets. Thus, the concentrations of species in a 0.35 M solution of Na2S04 can be expressed as follows [Na ] = 0.70 M and [SO ] = 0.35 M. Sample Problem 4.11 lets you practice relating concentrations of compounds and concentrations of individual ions using solution stoichiometry. 

Squate brackets around a chemical species can be read as the concentration of that spedes. For example, [NaT is read as the concentration of sodium ion.  

Using syuaie-brackci uoiatiou, express the concentration of (a) chloride ion in a solution that is 1.02 M in AICI3, (b) nitrate ion in a solution that is 0.451 M in Ca(N03)2, and (c) Na COs in a 

Strategy Use the concentration given in each case and the stoichiometiy indicated in the corresponding chemical formula to determine the concentration of the specified ion or compound. 

Sample Problem 4.13 (page 152) lets yon practice relating concentrations of compoimds and concentrations of individual ions using solution stoichiometry. 

13 Given the quantity of any species participating in a chemical reaction for which the equation can be written, find the quantity of any other species, either quantity being measured in (a) grams, (b) volume of solution at specified molarity, or (c) (if gases have been studied) volume of gas at given temperature and pressure. 

Convert the moles of given species to moles of wanted species. 

In Section 10.2, the quantities of given and wanted species in Steps 1 and 3 were measured in grams. In Chapter 14, either or both quantities were measured in (a) grams or (b) volume of gas at a known temperature and pressure. In this section, either or both quantities may be measured in (a) grams or (b) volume of solution of known concentration. 

Equation 10.2, the unit path for a problem for mass only, was 

In Chapter 3 we studied stoiehiometric calculations in terms of the mole method, which treats the eoeffieients in a balanced equation as the number of moles of reactants and products. In working with solutions of known molarity, we have to use the relationship MV = moles of solute. We will examine two types of common solution stoichiometry here gravimetric analysis and acid-base titration. 

Gravimetric analysis is an analytical technique based on the measurement of mass. One type of gravimetric analysis experiment involves the formation, isolation, and mass determination of a precipitate. Generally, this proeedure is applied to ionie eom-pounds. A sample substance of unknown composition is dissolved in water and allowed to react with another substanee to form a precipitate. The precipitate is filtered off, dried, and weighed. Knowing the mass and chemical formula of the precipitate formed, we can calculate the mass of a partieular chemical component (that is, the anion or cation) of the original sample. From the mass of the component and 

A reaction that is often studied in gravimetric analysis, because the reactants can be obtained in pure form, is 

Gravimetric analysis is a highly accurate technique, because the mass of a sample can be measured accurately. However, this procedure is applicable only to reactions that go to completion, or have nearly 100 percent yield. Thus, if AgCl were slightly soluble instead of being insoluble, it would not be possible to remove all the CP ions from the NaCl solution and the subsequent calculation would be in error. 

5662-g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNOs. If 1.0882 g of AgCl precipitate forms, what is the percent by mass of Cl in the original compound  

In the Check Y)ur Understanding of Example Problem 4.6, you considered a reaction to produce diborane. An alternative route to this propellant is shown below  

If 173.2 g of BF3 react with excess NaBHt to give 28.6 g of B2H6, what is the percentage yield  

In the stoichiometry problems we have discussed so far, the numbers of moles involved have been obtained by using the ratio established by the molar mass of the substances. Now we can expand the range of problems that we can consider by using Equation 3.2 to determine the number of moles of substances when volumes are measured rather than masses. The heart of the stoichiometry problem, 

The only complication added by considering reactions in solution is the need to relate three variables, using Equation 3.2, rather than two variables using the molar mass ratio. Example Problem 4.9 shows how to approach these solution stoichiometry problems. 

As we mentioned in Example Problem 3.10, the fuel hydrazine can be produced by the reaction of solutions of sodium hypochlorite and ammonia. The relevant chemical equation is 

In Section 4.2 we discussed how the coefficients in chemical equations are used as conversion factors between the amounts of reactants (in moles) and the amounts of products (in moles). In aqueous reactions, quantities of reactants and products are often specified 

We make the conversions between solution volumes and amounts of solute in moles using the molarities of the solutions. We make the conversions between amounts in moles of A and B using the stoichiometric coefficients from the balanced chemical equation. Example 4.8 demonstrates solution stoichiometry. 

What volume (in L) of 0.150 M KCl solution will completely react with 0.150 L of a 0.175 M Pb(N03)2 solution according to the following balanced chemical equation  

SORT You are given the volume and concentration of a Pb(N03)2 solution. You are asked to find the volume of KCl solution (of a given concentration) required to react with it. 

STRATEGIZE The conceptual plan has the form volume A amount A (in moles) amount B (in moles) — volume B. Use the molar concentrations of the KCl and Pb(N03)2 solutions as conversion factors between the number of moles of reactants in these solutions and their volumes. Use the stoichiometric coefficients from the balanced equation to convert between number of moles of Pb(N03)2 and number of moles of KCl. 

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In Chapter 4, molarity was the concentration unit of choice in dealing with solution stoichiometry. You will recall that molarity is defined as... 

We could again apply the seven-step process in detail. Instead, we take a more compact approach. Begin by determining what species are present in the reaction mixture. Next, use the solubility guidelines to identify the precipitate. After writing the balanced net ionic reaction, use solution stoichiometry and a table of amounts to find the required quantities. 

Reaction stoichiometry can then be used to solve for the molarity of the acid solution. See the Stoichiometry chapter for a discussion of solution stoichiometry. 

We discuss solutions further in the chapter on solutions and colligative properties, but solution stoichiometry is so common on the AP exam that we will discuss it here briefly also. Solutions are homogeneous mixtures composed of a solute (substance present in smaller amount) and a solvent (substance present in larger amount). If sodium chloride is dissolved in water, the NaCl is the solute and the water the solvent. 

As well as the solid solution formula given above, there have been suggestions that compositions off the join may also give single-phase NASICON. Part of the problem in determining solid solution stoichiometries and limits in materials such as NASICON arises because of the... 

Figure 6.4 Experimental curves for equilibrium gel filtration, (a) The 1-mL tuberculin syringe was incubated in 109-pAf[14C]valine (O) or 109-/zM[14C]valine and 4-mM ATP ( ). Then 100 fjL of a solution of 26-fiM valyl-tRNA synthetase was added to the same solution. Stoichiometries of 0.8 and 1.1, respectively, were found for the binding of the amino acid. Note the return to baseline between the peak and the trough— the mark of a good equilibrium gel filtration experiment, (b) An artifact-induced double peak obtained from the binding of [y-32P]ATP and valine to the enzyme. Some of the labeled ATP hydrolyzed to [32P]orthophosphate, which traveled down the column faster than the fy-32P]ATP did.

The relationship v/M XVj = MfX Vfis convenient for dilutions only. Students tend to use it for solution stoichiometry problems, which only works if file stoichiometry is 1 1. 

Advantages of volumetric standard solutions Solution stoichiometry... 

Chapter 4 Types of Chemical Reactions and Solution Stoichiometry... 

Molarity M mol solute L solution in solution stoichiometry calculations... 

Calculating Parts per Million Sample Problem A p. 461 Preparing 1.000 L of a 0.5000 M Solution Skills Toolkit 1 p.463 Calculating Molarity Skills Toolkit 2 p.464 Sample Problem B p. 465 Solution Stoichiometry Sample Problem C p. 466... 

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