Equation mole ratios

Chemical Formulas Balancing Chemical Equations Mole Ratios... 

To solve the problem, you need to know how the unknown moles of hydrogen are related to the known moles of potassium. In Section 11.1, you learned to derive mole ratios from the balanced chemical equation. Mole ratios are used as conversion factors to convert the known number of moles of one substance to the unknown number of moles of another substance in the same reaction. Several mole ratios can be written from the equation, but how do you choose the correct one ... 

Why learn to write mole ratios They are the key to calculations that are based on chemical equations. Using a balanced chemical equation, mole ratios derived from the equation, and a given amount of one of the reactants or products, you can calculate the amount of any other participant in the reaction. 

In a study of the kinetics of the reaction of 1-butanol with acetic acid at 0—120°C, an empirical equation was developed that permits estimation of the value of the rate constant with a deviation of 15.3% from the molar ratio of reactants, catalyst concentration, and temperature (30). This study was conducted usiag sulfuric acid as catalyst with a mole ratio of 1-butanol to acetic acid of 3 19.6, and a catalyst concentration of 0—0.14 wt %. 

Strategy First (1), convert the masses of the three elements to moles. Knowing the number of moles (n) of K, Cr, and O, you can then (2) calculate the mole ratios. Finally (3), equate the mole ratio to the atom ratio, which gives you the simplest formula. 

The law of combining volumes, like so many relationships involving gases, is readily explained by the ideal gas law. At constant temperature and pressure, volume is directly proportional to number of moles (V = kin). It follows that for gaseous species involved in reactions, the volume ratio must be the same as the mole ratio given by the coefficients of the balanced equation. 

From these and other related equations, it can be seen that, by using internal programs based on phosphate blends with specific sodium-to-phosphate (Na PQ4) mole ratios, BW alkalinity may be controlled... 

The balanced chemical equation for a reaction is used to set up the mole ratio, a factor that is used to convert the amount of one substance into the amount of another. 

To find out the mass of a product that can be formed from a known mass of a reactant, we first convert the grams of reactant into moles, use the mole ratio from the balanced equation, and then convert the moles of product formed into grams. Essentially, we go through three steps ... 

Step 2 Use the mole ratio derived from the stoichiometric coefficients in the balanced chemical equation to convert from the amount of one substance (A) into the amount in moles of the other substance (B). For aA - / B or aA + hY> — cC, use... 

Step 2 (a) Write the chemical equation for the reaction, (b) infer the mole ratio between the titrant species and the analyte species and (c) use it to convert the amount of titrant into amount of analyte (rValvte)-... 

Step 2 (a) Write the chemical equation, (b) Infer the mole ratio. 

In this approach, we use the mole ratio from the chemical equation to determine whether there is enough of one reactant to react with another. 

The poly(glycolide-co-caprolactone) (PGCL) copolymer was mainly synthesized by the ringopening polymerization. A copolymer with 1 1 mole ratio was synthesized by the ring-opening polymerization in the presence of the catalyst Sn(Oct)2 by Lee and coworkers. The polymerization was under vacuum, and heated in an oil bath at 170°C for 20 h. The copolymer was then dried under vacuum at room temperature for 72 h. The schematic reaction equations are shown in Schemes 8.5 and 8.6. 

C04-0146. The largest single use of sulfuric acid is for the production of phosphate fertilizers. The acid reacts with calcium phosphate in a 2 1 mole ratio to give calcium sulfate and calcium dihydrogen phosphate. The mixture is crushed and spread on fields, where the salts dissolve in rain water. (Calcium phosphate, commonly found in phosphate rock, is too insoluble to be a direct source of phosphate for plants.) (a) Write a balanced equation for the reaction of sulfuric acid with calcium phosphate, (b) How many kilograms each of sulliiric acid and calcium phosphate are required to produce 50.0 kg of the calcium sulfate-dihydrogen phosphate mixture (c) How many moles of phosphate ion will this mixture provide ... 

These above equations suggest that amines with pKa In the range of 4-6 will be more rapidly nltrosated than those with pKa values In the range of 9-11. This has been borne out In practice many times. Amines such as N-methylanlllne with a pKa value of 4.84, piperazine (pKa value 5.9, 9.8), and amlno-pyrlne (pKa value 5.04) are much more rapidly nltrosated than plperdlne (pKa value 11.2), dlmethylamlne (pKa value 10.72), and pyrrolidine (pKa value 11.27). Under many of the conditions studied. It has been shown that the reactivity of ascorbic acid Is sufficiently rapid that It can successfully compete with most all amines when present In approximately 2 mole ratio excess of the nltrosatlng agent. 

Fig. 4.18 represents a countercurrent-flow, packed gas absorption column, in which the absorption of solute is accompanied by the evolution of heat. In order to treat the case of concentrated gas and liquid streams, in which the total flow rates of both gas and liquid vary throughout the column, the solute concentrations in the gas and liquid are defined in terms of mole ratio units and related to the molar flow rates of solute free gas and liquid respectively, as discussed previously in Sec. 3.3.2. By convention, the mass transfer rate equation is however expressed in terms of mole fraction units. In Fig. 4.18, Gm is the molar flow of solute free gas (kmol/m s), is the molar flow of solute free liquid (kmol/m s), where both and Gm remain constant throughout the column. Y is the mole ratio of solute in the gas phase (kmol of solute/kmol of solute free gas), X is the mole ratio of solute in the liquid phase (kmol of...

The gas is at high concentration and therefore the column component balance equations are based on mole ratio concentration units. The form of the balance equations follow those of Sec. 4.4.1. 

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. 

Mole ratios can be used to determine the amount of one substance needed to react with a given amount of another substance. In this experiment, you will react a substance called an acid with another substance called a base. Acids can be defined as substances that dissociate and produce hydrogen (H+) ions when dissolved in water. Bases are substances that ionize to produce hydroxide (OH ) ions when they dissolve in water. When acids and bases react with each other, the H+ ions and OH ions join to form water (H20). The resulting solution no longer has an excess of either H+ ions or OH- ions. The solution has become neutral. This process is called neutralization. By using the mole ratio of hydrogen ions and hydroxide ions in the balanced chemical equation, you can predict the point at which a solution becomes neutral. 

A chemical equation describes a chemical reaction in many ways as an empirical formula describes a chemical compound. The equation describes not only which substances react, but the relative number of moles of each undergoing reaction and the relative number of moles of each product formed. Note especially that it is the mole ratios in which the substances react, not how much is present, that the equation describes. In order to show the quantitative relationships, the equation must be balanced. That is, it must have the same number of atoms of each element used up and produced (except for special equations that describe nuclear reactions). The law of conservation of mass is thus obeyed, and also the "law of conservation of atoms. Coefficients are used before the formulas for elements and compounds to tell how many formula units of that substance are involved in the reaction. A coefficient does not imply any chemical bonding between units of the substance it is placed before. The number of atoms involved in each formula unit is multiplied by the coefficient to get the total number of atoms of each element involved. Later, when equations with individual ions are written (Chap. 9), the net charge on each side of the equation, as well as the numbers of atoms of each element, must be the same to have a balanced equation. The absence of a coefficient in a balanced equation implies a coefficient of 1. 

Since the mole ratio is given by the equation, we must convert the 100.0 g of NaOH to moles ... 

Since the equation states the mole ratio, wc first convert the number of molecules of H3P04 to moles ... 

Using these expressions and Equation 19.2, and recognizing that the mole ratio ni60/ o of 160 to elemental oxygen is nearly constant, we can show that the total (or bulk) isotopic composition of the isotope system,... 

Another approach is to define concentration in the micellar pseudophase in terms of a mole ratio. Concentration is then defined unambiguously, and the equations take a simple form (Bunton, 1979a,b Romsted, 1984). However, this approach does not allow direct comparison of second-order rate constants in aqueous and micellar pseudophases and by evading one problem one faces another. 

B The assumptions include no heat loss to the surroundings or to the calorimeter, a solution density of 1.00 g/mL, a specific heat of 4.18 J g 1 °C 1, and that the initial and final solution volumes are the same. The equation for the reaction that occurs is NaOH(aq) + HCl(aq) - NaCl(aq) + H20(l). Since the two reactants combine in a one to one mole ratio, the limiting reactant is the one present in smaller amount. 

Ethyl formate is to be produced from ethanol and formic acid in a continuous flow tubular reactor operated at a constant temperature of 303 K (30°C). The reactants will be fed to the reactor in the proportions 1 mole HCOOH 5 moles C2H5OH at a combined flowrate of 0.0002 m3/s (0.72 m3/h). The reaction will be catalysed by a small amount of sulphuric acid. At the temperature, mole ratio, and catalyst concentration to be used, the rate equation determined from small-scale batch experiments has been found to be ... 

The mole is the most important concept in this chapter. Nearly every problem associated with this material requires moles in at least one of the steps. You should get into the habit of automatically looking for moles. There are several ways of finding the moles of a substance. You may determine the moles of a substance from a balanced chemical equation. You may determine moles from the mass and molecular weight of a substance. You may determine moles from the number of particles and Avogadro s number. You may find moles from the moles of another substance and a mole ratio. Later in this book, you will find even more ways to determine moles. In some cases, you will be finished when you find moles, in other cases, finding moles is only one of the steps in a longer problem. 

The term (2 mol HC1/1 mol H2) is a mole ratio. We got this mole ratio directly from the balanced chemical equation. The balanced chemical equation has a 2 in front of the HC1, thus we have the same number in front of the mol HC1. The balanced chemical equation has an understood 1 in front of the H2, for this reason the same value belongs in front of the mol H2. The values in the mole ratio are exact numbers, and, as such, do not affect the significant figures. 

To find the moles of IF5 from the limiting reagent, we need to use a mole ratio derived from information in the balanced chemical equation. (This is another place where, if we had not balanced the equation, we would be in trouble.)... 

Now let us try an example needing additional information after the mole ratio step. I low many grams of calcium hydroxide are necessary to titrate 0.200 mol of acetic acid As usual, we begin by adding this information to the balanced chemical equation ... 

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