The Extent of Reaction equation can be used to specify states of matter or reaction conditions.
The method of converting to, between, and from moles can be used to solve the problems.
Determine the molarity of a solution from its measured quantities, and determine the volume of solution used in a solution dilution or a chemical reaction.
Define the terms limiting and excess reactant, and describe how to determine which reactant is the limiting one in a chemical reaction.
Determine the theoretical and percent yield of a reaction.
The space shuttle Discovery is about to lift off on a mission.
In this chapter, we learn how to use balanced chemical equations for a wide variety of chemical reactions.
We know that iron rusts and natural gas burns.
These are chemical reactions.
The entire science of chemistry is concerned with chemical reactions.
In describing chemical reactions, we tend to focus on the atoms, ion, or molecule that make up the substances involved.
Chemical Reactions quantities of substances can be easily measured or manipulated.
The relationships we need to relate macroscopic amounts of substances to our microscopic view of chemical reactions are provided by reaction stoichiometry.
Stoichiometry is important, even if it is not more exciting than the law of mass.
In order to make new discoveries and expand our knowledge of the world of atoms, molecules, and ion, chemists use the principles of stoichiometrics to plan experiments, analyze their results, and make predictions.
In this chapter, we will learn to represent chemical reactions by Chemi cal equations, and then we will use chemical equations and ideas from earlier chapters to establish the quantitative relationships we seek.
We will discuss new aspects of problem solving in the chapter.
A chemical reaction is the process by which a chemical change occurs.
When substances are mixed, they retain their original composition and properties.
Evidence is needed before we can say that a reaction occurred.
A detailed chemical analysis of the silver nitrate and potassium reaction mixture is needed to identify all the substances present.
The analysis may show that a chemical reaction has occurred even in the absence of obvious physical signs.
The left side of the equation and the formulas for the products are written on the physical evidence of a right.
We said that reaction.
To get the following expression, substitute chemical formulas for names.
There are three O atoms on the left side, one in the mol reactions and two in the molecule O ward direction.
The molecule NO has two O atoms on the right.
There are no atoms in Chapter 15.
To get a balanced chemical equation, balance the numbers of atoms on both sides of the expression.
The reaction of NO and O2 is placed in front of the formulas NO and NO2.
Two mole is consumed and two molecules of NO mental interest.
Every molecule of O engines is produced for automobile 2.
There are two N atoms and four O atoms on each side of the equation.
We can see this in the symbolic equation and in the amount of NO emitted from a representation of the reaction.
The reaction illustrated on this page is 2 NO2 brown NO2 Many urban atmospheres have a brown tint.
The amounts of reactants used and products formed in a chemical reaction are related through a variety of calculations.
Keep in mind the following point when balancing a chemical equation.
Adjusting the coefficients of formulas is the only way to balance an equation.
By inspection, balancing means adjusting coefficients by trial and error until a balanced condition is found.
The elements can be balanced in any order, but not always.
There are some useful strategies for balancing equations.
If there is only one compound on each side of the equation, try balancing this element first.
Balance the free element when one of the reactants or products is present.
Some groups of atoms remain the same in some reactions.
The left and right sides must be equal in an equation.
An equation should not be called an expression until it is balanced.
In Chapters 7 and cients, it is permissible to use fractional and integral numbers.
You should learn to balance the equation and write formulas for reactants and products of a reaction in this chapter.
You will face a third task, which you will face in later chapters, which is to make sure that the strategies described in the book work well for simple conditions.
Based on pre reactions.
In a plentiful supply of oxygen gas, car methods must be employed.
Sulfur dioxide will be a product as well.
Writing and balancing an equation is used in the plasticizer industry.
There is a plentiful supply of oxygen and a balanced chemical equation is needed for the combustion of this compound.
A ball-and-stick model of triethylene glycol is shown.
The formula of triethylene glycol is deduced from the model.
There are 6 C atoms, 4 O atoms, and 14 H atoms.
The formula is C H O.
After identifying the reactants and products, we balance the expression with respect to each kind of atom by writing down an unbalanced chemical expression for the reaction.
There are 19 O atoms on the right side of the expression and only 4 on the left side.
15/2 for O2 is needed to get 15 more O atoms.
The numbers of C, H, and O atoms on each side of the equation are used to check if the equation is balanced.
A balanced equation is needed to represent the reaction of mercury(II) sulfide and calcium oxide to produce calcium sulfide, calcium sulfate, and mercury metal.
A balanced equation is needed for the combustion of thiosalicylic acid, C7H6O2S, used in the manufacture of indigo dyes.
There is a balland-stick model of C Thiosalicylic acid.
If our interest is only in balancing an equation, these facts are meaningless.
Sometimes it is necessary to include this information in a chemical equation in order to convey a complete representation of the reaction.
The physical form of reactants and products is shown in parentheses.
There is not enough information in the equation for a chemical reaction to be carried out in a laboratory or chemical plant.
Working out the conditions for a reaction is an important part of modern chemical research.
Above or below the arrow is where the reaction conditions are written.
The Greek capital letter C/ means that a high temperature is required for the reaction mixture to be heated.
A catalyst is a substance that enters into a reaction in such a way that it speeds up the reaction without itself being consumed or changed by the reaction.
When certain quantities of reactants are consumed, it is important to calculate how much of a particular product will be produced.
We will see how to use chemical equations to set up conversion factors in the next section.
The requirement that the number of atoms be equal on either side of the equation is met is for elements K, Cl, and O.
In Chapter 3, we looked at the quantitative meaning of chemical formulas.
2x molecule H2O, let x be 6.02214 * 1023
We can turn such statements into conversion factors.
The factors are printed in blue.
There is more than enough O2 available to allow the complete conversion of the triethylene glycol to CO2 and H2O.
The equation for the combustion reaction has a factor for converting moles of C6H14O4 to moles of CO2
A balanced equation is the first step in a calculation.
The equation for the reaction is given below.
12 mol CO2 are produced for every 2 mol C6H14O4 burned.
The production of 12 mol CO2 is equivalent to the consumption of 2 mol C6H14O4.
It is easier to set up an expression in terms of ratios and then solve it for an unknown quantity.
The appropriate mole ratio that converts from moles A to moles B is a key step in working a stoichiometric problem.
It is important to include Figure 4-3 in this strategy because it will yield information about one substance, B, from information given units and to work from a bal about a second substance, A.
The conversions can be combined into a single-step calculation, as shown in the following examples.
The general strategy involves three conversions: (1) to moles, (2) between moles and (3) from moles.
The required conversions are g C6H14O4 1 " mol C6H14O4 2 " mol CO2 3 " g CO2.
The conversion factor changes the unit on the left to the one on the right.
Either a stepwise approach or a conversion pathway approach can be used to carry out the conversions.
There is a connection between the stepwise approach and the conversion pathway approach.
The second conversion factor converts moles of C6H14O4 to moles of CO2.
The number to the right of 4.16 g C6H14O4 has a value of 6 and the number between 250 and 300 has a factor of less than 2.
The mass of CO2 should be less than that of the C6H14O4.
In the ultimate conversion from g C6H14O4 to g CO2, all units cancel properly.
The required conversions are g C6H14O4 1 and mol C6H14O4 2.
We will use a stepwise approach to solve the problem.
The factor was used by 2.
There is a single line calculation shown.
The number to the right of 6.86 g C6H14O4 has a value of 7.5 and the product is about 32.00, which is a factor of about 250.
The mass of O2 should be about 5>3 that of C6H14O4 and it is less than 12.
Other common conversions may require factors such as volume, density, and percent composition.
We must always use the appropriate factor from the chemical equation as a key conversion factor.
Small volumes of hydrogen gas, H2(g), can be prepared in the laboratory using the reaction between solid aluminum, Al(s), and hydrochloric acid.
The equation for the reaction is shown.
There is a simple laboratory setup for collecting hydrogen gas.
A range of possibilities for calculations can be found in the reaction between Al and HCl.
There is a flask on the left.
The liberated H2(g) flows into a gas-collection apparatus.
Water has hydrogen in it.
The alloy used in aircraft structures is 93.7% Al and 6.3% Cu.
The density of the alloy is 2.65 g> cm3.
There is a simple approach to this calculation outlined.
The conversion factor changes the unit on the left to the one on the right.
A single setup in which the five conversions are performed in sequence can be used for the calculation.
We use a stepwise approach.
The factor is used to calculate H2 by using it.
Store intermediate results without rounding them off.
We don't have to write down intermediate results when we combine the steps into a single calculation.
We have to evaluate whether the answer is a reasonable one.
The mass of Al and H2 is 27 g/mol and 2 g/mol, respectively.
The equation tells us that 1 mole of Al produces 1.5 mol of H2 and that it weighs 1.5 * 2.
We expect less than 0.3 g of H2 because we are dealing with less than 2.7 g of Al.
The answer is reasonable.
A hydrochloric acid solution consists of 28.0% HCl by mass and has a density of 1.14 g>mL.
The first challenge is to figure out where to start.
The starting point for the problem is with the given information--1.87 g Al.
The goal of the calculation is a solution volume.
We use a stepwise approach.
If you want to convert grams of Al to moles of 1 mol Al, use the molar mass of Al.
The convert from moles of Al to moles of 6 mol HCl is done using the stoichiometric factor.
The mass of HCl is used.
The individual steps are combined into a single line.
Assess Let's try to figure out if the answer is reasonable by using numbers that are rounded off slightly.
The density of the solution is 1 g/mL and the solution is 30% HCl by mass, so a 24 mL sample of the solution will contain approximately 0.30 grams of HCl.
The equation tells us that 1mol Al, or 27 g Al, reacts with 3mol HCl.
Stated another way, 4 g HCl reacts with 1 g Al.
In this example, we are using approximately 7.2 g HCl; thus, we consume approximately 7.2 g HCl.
This value is close to the actual amount of Al consumed and we conclude that our answer is reasonable.
A particular vinegar has a concentration of 4% CH3COOH by mass.
It reacts with the carbonate to produce gasses.
The density of the liquid is 1.01 g.
The maximum quantity of O2(g) per gram of reactant can be determined without performing detailed calculations.
The majority of reactions are carried out in solution.
The close contact between atoms, ion, or molecule necessary for a reaction to occur is achieved by mixing the reactants in solution.
There are a few new ideas that are helpful.
Liquid water is the solvent.
To describe a solution in which liquid water is the solvent and NaCl is the solute, we use the notation NaCl(aq).
A solution with 0.440mol CO(NH2)2>L is 0.440 M CO(NH2)2 or 0.440 molar CO(NH2).
If 0.110 mol urea is present in 250.0 mL of solution, the solu tion is also 0.440 M.
When calculating concentration, we must determine the amount of solute in moles from other quantities that can be easily measured, such as the mass of solute or the volume of a liquid solute.
The solution volume is converted to liters.
The method commonly used to prepare a solution is not depicted in Figure 4-5.
A used, or the sum of the solid sample is weighed out and dissolved in sufficient water to produce a volumes of solvent and solution.
It is not solute to fill a beaker to a 250 mL mark.
If we add nearly precise enough as a volume measurement.
The beaker's solution volume would be a systematic error because it wouldn't be 250.0 mL nor enough precision.
In Chapter 14 we learn how to bring the solution level to the 250 mL mark.
The graduated cylinder is more precise than the beaker, but that doesn't mean the error won't be more.
The number of moles of solute in a carefully measured volume of solution can be calculated using the equation below.
A solution is prepared by dissolving 25.0 mL of alcohol, CH3CH2OH, in enough water to produce 250.0 mL solution.
We need to know how many moles are in the sample.
The solution's molarity is calculated using an equation.
The number of moles in the sample is calculated in a single line.
To apply the definition of molarity, note that 250.0 mL is 0.2500 L.
It's important to include the units in this calculation to make sure we get the correct units for the final answer.
When dealing with liquid solutes, be careful to distinguish between the two.
A sample of acetone is dissolved in enough water to produce a solution of 1.25 liters.
Adding 12.1 g K2CrO4(s) to 250.0 mL water is not enough to prepare the solution.
The weighed quantity of K2CrO4(s) is first added to a clean, dry 250 mL volumetric flask.
L soln is the conversion pathway.
The first conversion factor is the molarity of the solution, and the second is the molar mass.
The answer has the correct units.
We can use numbers that are rounded off to see if the answer is reasonable.
The number of moles of K2CrO4 in the sample is approximately 12/200, since the mass of the sample is approximately 12 g. 0.24 mol/L is the approximate molarity.
We are confident that the answer, 12.1 g K2CrO4, is correct because it is close to the true molarity.
A solution saturated with NaNO3 is 10.8 M.
Rows of bottles containing solutions for use in chemical reactions are a common sight in chemistry storerooms and laboratories.
It's not practical to keep solutions of every concentration.
If we write equation (4.4) for the initial solution, we get ni, and if we write equation (f) for the final solution, we get nf.
The laboratory procedure for preparing a solution by dilution is shown in Figure 4-7.
The equation is only applicable to dilution problems.
There are 2 moles of substance 1 and 2 moles of substance 2.
A particular analytical chemistry procedure is required.
The number of moles that must be present in the final solution is calculated first.
The amount of K2CrO4 is 0.250 M.
The amount ofsolute must be calculated first.
The volume of 0.250 M K2CrO4 is equivalent to 250 mol K2CrO4.
The equation is an alternative approach.
The volume of solution to be prepared and the concentrations of the final and initial solutions are known factors.
We have to solve for the initial volume.
If we use the same unit for both Vf and Vi in the equation, any volume unit can be used.
The term needed to convert volumes to liters would appear on both sides of the equation.
If we add 0.250 M K2CrO4 to 0.250 L with water, the concentration of the solution is 0.010 L. The answer, 10.0 mL of 0.250 M K2CrO4, is correct.
A 15.00 mL sample of 0.450 M K2CrO4 is being tested.
When left in an open beaker for a period of time, the volume of 0.105 M NaCl is found to decrease to 237 mL.
The solution is 0.0100 M K2CrO4.
The use of molarity as a conversion factor from solution volume to number of moles of reactant is different from previous examples.
There are more examples of calcu lations involving solutions in Chapter 5.
An excess of AgNO3 is added to a pipetful of 0.250 M K2CrO4.
The excess of AgNO3 is used to tell us that all of the K2CrO4 is consumed.
A mass of Ag2CrO4 is expressed in grams at the end of the calculation.
Use a stepwise approach to solve this problem.
Combining the steps into a single line calculation can give you the same answer.
It is always a good sign that the units work out.
Let's work the problem in reverse and use numbers that are rounded off slightly.
A sample of Ag2CrO4 has 1/332 moles.
The K2CrO4 solution has a molarity of 0.24 M and the number of moles in the sample is approximately 0.
Sometimes this condition is needed, for example, in a certain chemical reactant.
Sometimes one of the reactants is consumed in a precipitation reaction.
The limiting reactant may not be called explicitly in some cases.
Sometimes our interest in a limiting reactant problem is in determining how much of an excess reactant remains, as well as how much product is formed.
PCl3 is a commercially important compound used in the manufacture of pesticides, gasoline Additives, and a number of other products.
There is a ball-and-stick model of PCl3.
Liquid PCl3 is made from a combination of chlorine and phosphorus.
Correctly identifying the limiting reactant is the key to solving this problem.
If more than 6 mol Cl2 is available per mole, chlorine is in excess and P4 is the lim Phosphorus trichloride iting reactant.
If less than 6 mol Cl2 is available per mole, chlorine is the limiting reactant.
Chlorine is the limiting reactant since there is less than 6mol of it per mole.
The mass of PCl3 will be determined by the remainder of the calculation.
We can complete the calculation using a stepwise approach after we identified the limiting reactant.
We combine the steps into a single line calculation using the conversion pathway approach.
There is a different approach that leads to the same result.
First, calculate the mass of PCl3 produced by the reaction of 323 g P4 with an excess of P4 and then the mass of PCl3 produced by the reaction of 125 g P4.
This approach can be generalized to deal with many reactants.
We can calculate the mass of P4 using either the mass of PCl3 produced or the mass of Cl2 consumed.
The following conversion pathway is used to calculate the mass of P4 that is consumed.
The following stepwise approach can be used to calculate the mass of P4.
The following is the single line calculation.
The answer is obtained from the mass of PCl3 that is produced.
It is possible to use the law of mass to make sure the result is correct.
The total mass of the reactants must be equal to the mass of the product.
The mass of the remains is 448 g P4 417 g, which is 31 g. There are no digits after the decimal point in 125.
If the reaction of 1.0 mol NH3(g) and 1.0 mol O2(g) is carried to completion, it will produce NO(g) and H2O(l) as the only products.
In this section, we look at a few more factors in the manufacturing plant and in the laboratory.
The calculated outcome of a reaction may not be what is actually observed.
The amount of product may be less than expected.
The route to producing a desired chemical may require several reactions in sequence.
Two or more reactions may occur at the same time.
Quantitative chemical analyses can be done with such reactions.
The actual yield is less than the theoretical yield, and the percent yield is less than 100%.
There may be a variety of reasons for the reduced yield.
The yield of the main product is reduced if side reactions occur.
The apparent yield can be greater than 100% at times.
This situation indicates an error in technique because we can't get anything from nothing.