Reactant Problems

To solve a quantitative limiting reactant problem, we identify the limiting reactant by working with amounts in moles and the stoichiometric coefficients from the balanced equation. For the ammonia synthesis, if we start with 84.0 g of molecular nitrogen and 24.2 g of molecular hydrogen, what mass of ammonia can be prepared First, convert from... 

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. 

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. 

Data are given for all reactants, so this is a limiting reactant problem. We must balance the chemical equation and then work with a table of molar amounts. 

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

Balanced equation 2CO + 02 -> 2C02 This is a limiting reactant problem. 

However, this is a limiting reactant problem with propane, C3H8, being the limiting reactant since we are given 7 molecules of 02 and not 5. When the reaction is complete, there will be 2 molecules of 02 remaining unreacted. 

We do not have stoichiometric amounts of both reactants this is a limiting reactant problem. We have less H3P04 than is necessary to react with all of the NaOH, so H3P04 is the limiting reactant and NaOH is in excess. The amount of salt formed is set then by the amount of H3P04. 

This is a possible limiting reactant problem because amounts of both reactants are given. In this case, it is clear that Mg is the limiting reactant, since... 

New NH3/NH4+ buffer When 0.142 mol per liter of HC1 is added to the original buffer presented in (a), it reacts with the base component of the buffer, NH3, to form more of the acid component, NH4+ (the conjugate acid of NH3). Since HC1 is in the gaseous phase, there is no total volume change. A new buffer solution is created with a slightly more acidic pH. In this type of problem, always perform the acid-base limiting reactant problem first, then the equilibrium calculation. 

Plan (1) Perform the acid-base neutralization limiting reactant problem. 

For the rest of the exercise, the plan is straightforward for the neutralization reaction between HN03 and NaOH, perform the limiting reactant problem. The pH is determined from the concentration of excess HN03 or NaOH. Each calculation is totally independent of the other calculations. 

Before the equivalence point, the pH is determined by the buffer solution consisting of the unreacted CH3COOH and NaCH3COO produced by the reaction. Each calculation is a limiting reactant problem using the original concentration of CH3COOH. 

NH4+, derived from the salt is negligible. For example, consider the limiting reactant problem for point (c)... 

Plan (1) Do the limiting reactant problem to determine which reactant is in excess. 

Compute the amount of NH3 formed from each reactant in this limiting reactant problem. 

This is similar to a limiting reactant problem. We determine how many necklaces can be made from each quantity of beads. 

Next, realize that this problem is a limiting-reactant problem. That is, one of the two reactants will run out first, and when that happens, the reaction will stop. You need to determine which one of the reactants will run out first. To do this, you need to be able to compare them on a 1 1 basis. But their coefficients are different, so you need to relate both reactants to a common product, say Ba3(As04)2. Set the problem up like this ... 

Stoichiometry problems (including limiting-reactant problems) involving solutions can be worked in the same fashion as before, except that the volume and molarity of the solution must first be converted to moles. 

In limiting-reactant problems, don t consider just the number of grams or even moles to determine the limiting reactant—use the mol/coefficient ratio. 

Begin by realizing that this is a limiting-reactant problem (yon may run out of either Mg(OH)2 or HCl). Both reactants must be considered in how much MgCl2 they can produce. This problem can be done throngh the factor-label method. 

From Pair Kinetics Toward the Many-Reactant Problem... 

From pair kinetics toward the many-reactant problem... 

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