Limiting reactant calculations

Note that we solved this problem by first performing a stoichiometric (limiting reactant) calculation and then an equilibrium calculation. A similar strategy works if a strong base such as OH is added instead of a strong acid. The base reacts with formic acid to produce formate ions. Adding 0.10 mol of OH to the HCOOH/HCOO buffer of Example 15.7 increases the pH only to 3.58. In the absence of the buffer system, the same base would raise the pH to 13.00. 

Problem-Solving Tip Review Limiting Reactant Calculations... 

There is another method that some of our students find works well They calculate the mass of product expected based on each reactant. The limiting reactant is that reactant that gives the smallest quantity of product. For example, refer to the SiCL reaction with Mg on page 158. To confirm that SiCL is the limiting reactant, calculate the quantity of elemental silicon that can be formed starting with (a) 1.32 mol of SiCL and unlimited magnesium or (b) with 9.26 mol of Mg and unlimited SiCL. 

Referring to the reactants as A and B, to determine if reactant A is the limiting reactant, calculate the number of moles of B it will consume (a mole-to-mole conversion). 

Check if Al is the limiting reactant. Calculate the moles of S consumed by 6.00 moles of Al. 

N9 was the limiting reactant. Calculate how much H9 is left over. 

In one experiment 0.886 mole of NO is mixed with 0.503 mole of O2. Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced. 

If 0.740 g of O3 reacts with 0.670 g of NO, how many grams of NO2 will be produced Which compound is the limiting reactant Calculate the number of moles of the excess reactant remaining at the end of the reaction. 

Cumene is produced from the reaction of benzene and propylene at 25°C and 1 atm. The inlet mass flow rate of benzene is 1000 kg/h and that of propylene is 180 kg/h. Assume 45% completion of the limiting reactant. Calculate molar flow rates of the product stream. 

In cases where the reactants involved are not present in the proper stoichiometric ratios, the limiting reactant will have to be determined and the excess amounts of the other reactants calculated. It is safe to assume that unconsumed reactants and inert components exit with the products in their original forms. Consider the following example. 

To simplify calculations, but also by convention, the amount of excess reactant in a reaction is defined on the basis of the reaction going to completion for the limiting reactant. In the case of methane (CH4) burned with excess air, the volume of air needed to combust the methane is calculated as though there is complete combustion of the methane, converting it entirely to carbon dioxide and water. 

Often you will be given the amounts of two different reactants and asked to determine which is the limiting reactant, to calculate the theoretical yield of the product and to find how much of the excess reactant is unused. To do so, it helps to follow a systematic, four-step procedure. 

Choose the smaller of the two amounts calculated in (1) and (2). This is the theoretical yield of product the reactant that produces the smaller amount is the limiting reactant. The other reactant is in excess only part of it is consumed. 

The theoretical yield is the maximum amount of product that can be obtained. In calculating the theoretical yield, it is assumed that the limiting reactant is 100% converted to product. In the real world, that is unlikely to happen. Some of the limiting reactant may be consumed in competing reactions. Some of the product may be lost in separating it from the reaction mixture. For these and other reasons, the experimental yield is ordinarily less than the theoretical yield. Put another way, the percent yield is expected to be less than 100% ... 

After reaction, the activity of a 25.0-mL water sample is 745 counts per minute (cpm), caused by the presence of Tl+-204 ions. The activity of Tl-204 is 5.53 X 105 cpm per gram of thallium metal. Assuming that 02 is the limiting reactant in the above equation, calculate its concentration in moles per liter. 

In some cases, we must determine by calculation which is the limiting reactant. For example, from the equation... 

The limiting reactant is the reactant that will be completely used up. All other reactants are in excess. Because the limiting reactant is the one that limits the amounts of products that can be formed, the theoretical yield is calculated from the amount of the limiting reactant. 

Step 3 If the actual amount of the second reactant is greater than the amount needed (the value calculated in step 2), then the second reactant is present in excess in this case, the first reactant is the limiting reactant. If the actual amount of the second reactant is less than that calculated, then all of it will react so it is the limiting reactant and the first reactant is in excess. 

Calculate the theoretical molar yield of one of the products for each reactant separately, by using the procedure in Toolbox L.l. This method is a good one to use when there are more than two reactants. The reactant that would produce the smallest amount of product is the limiting reactant. 

Step 1 Convert the mass of each reactant into moles, if necessary, by using the molar masses of the substances. Step 2 Select one of the products. For each reactant, calculate how many moles of the product it can form. Step 3 The reactant that can produce the least amount of product is the limiting reactant. 

EXAMPLE M.3 Calculating percentage yield from a limiting reactant... 

STRATEGY First, the limiting reactant must be identified (Toolbox M.l). This limiting reactant determines the theoretical yield of the reaction, and so we use it to calculate the theoretical amount of product by Method 2 in Toolbox L.l. The percentage yield is the ratio of the mass produced to the theoretical mass times 100. Molar masses are j calculated using the information in the periodic table inside the front cover of this i book. 

J 2 Identify the limiting reactant of a reaction and use the limiting reactant to calculate the yield of a product and the... 

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 ... 

A reaction that is carried out under limiting reactant conditions nevertheless has a yield that generally will be less than 100%. The reasons why reactions yield less than the theoretical amounts, given in Section 4-1. apply to all reactions. When a reaction operates under limiting reactant conditions, we calculate the theoretical yield assuming that the limiting reactant will be completely consumed. We then determine the percent yield as described in Section 4A. Example shows how to do this. 

The problem asks for a yield, so we identify this as a yield problem. In addition, we recognize this as a limiting reactant situation because we are given the masses of both starting materials. First, identify the limiting reactant by working with moles and stoichiometric coefficients then carry out standard stoichiometry calculations to determine the theoretical amount that could form. A table of amounts helps organize these calculations. Calculate the percent yield from the theoretical amount and the actual amount formed. 

Mixing the two solutions will produce 2.50 X 10 mol of Fe (0H)3 precipitate, which is 2.67 g. The mixed solution contains Na cations and Cl anions, too, but we can ignore these spectator ions in our calculations. Notice that this precipitation reaction is treated just like other limiting reactant problems. Examples and further illustrate the application of general stoichiometric principles to precipitation reactions. 

Any of the types of problems discussed in Chapters 3 and 4 can involve gases. The strategy for doing stoichiometric calculations is the same whether the species involved are solids, liquids, or gases. In this chapter, we add the ideal gas equation to our equations for converting measured quantities into moles. Example is a limiting reactant problem that involves a gas. 

C05-0040. Summarize the strategy for calculating final pressures of gases in a limiting reactant problem. 

The quantitative treatment of a reaction equilibrium usually involves one of two things. Either the equilibrium constant must be computed from a knowledge of concentrations, or equilibrium concentrations must be determined from a knowledge of initial conditions and Kgq. In this section, we describe the basic reasoning and techniques needed to solve equilibrium problems. Stoichiometry plays a major role in equilibrium calculations, so you may want to review the techniques described in Chapter 4, particularly Section 4- on limiting reactants. 

Again, we use the standard approach to an equilibrium calculation. In this case the reaction is a precipitation, for which the equilibrium constant is quite large. Thus, taking the reaction to completion by applying limiting reactant stoichiometry is likely to be the appropriate approach to solving the problem. 

Calculate the theoretical yield of Fe, based on the limiting reactant. 

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