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Chapter 9: Problem 44

What does it mean to say a reactant is present "in excess" in a process? Can the limiting reactant be present in excess? Does the presence of an excess of a reactant affect the mass of products expected for a reaction?

Short Answer

Expert verified
A reactant is present "in excess" when there is more of it than necessary for the chemical reaction to go to completion, meaning that some of it will remain unreacted. The limiting reactant cannot be present in excess, as it controls the maximum amount of product formed and is fully consumed during the reaction. The presence of an excess reactant does not change the expected mass of products in a reaction, as the limiting reactant determines the maximum amount of product formed, and the excess reactant does not contribute to product formation.

Step by step solution

01

Definition of a Reactant in Excess

A reactant is said to be present "in excess" when there is more of it than necessary for the chemical reaction to go to completion, based on the stoichiometry of the reaction. In other words, there is more of the reactant than is needed to react with the other reactants, which means that at the end of the reaction, some amount of the excess reactant will remain unreacted.
02

Limiting Reactant Presence in Excess

The limiting reactant cannot be present in excess because, by definition, the limiting reactant controls the maximum amount of product that can be formed in the reaction. It is the reactant that is fully consumed during the reaction, meaning that it determines how far the reaction can progress. When the limiting reactant is used up, the reaction stops, even if other reactants still remain in excess.
03

Impact of Excess Reactant on Mass of Products

The presence of an excess reactant does not change the expected mass of products in a reaction. As discussed earlier, the limiting reactant determines the maximum amount of product that can be formed. The excess reactant remains unreacted and does not contribute to the formation of additional product, so the expected mass of products stays the same, regardless of the excess reactant's presence. For example, when using stoichiometry in chemical calculations, we only consider the proportions of the reactants according to their balanced chemical equations, and do not include the excess reactant in our calculations.

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Most popular questions from this chapter

The compound sodium thiosulfate pentahydrate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O},\) is important commercially to the photography business as "hypo," because it has the ability to dissolve unreacted silver salts from photographic film during development. Sodium thiosulfate pentahydrate can be produced by boiling elemental sulfur in an aqueous solution of sodium sulfite. $$\mathrm{S}_{8}(s)+\mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)$$ (unbalanced) What is the theoretical yield of sodium thiosulfate pentahydrate when \(3.25 \mathrm{g}\) of sulfur is boiled with 13.1 g of sodium sulfite? Sodium thiosulfate pentahydrate is very soluble in water. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only \(5.26 \mathrm{g}\) of the product?

Solutions of sodium hydroxide cannot be kept for very long because they absorb carbon dioxide from the air, forming sodium carbonate. The unbalanced equation is $$\mathrm{NaOH}(a q)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains \(5.00 \mathrm{g}\) of sodium hydroxide.

For each of the following balanced reactions, calculate how many moles of product would be produced by complete conversion of 0.15 mol of the reactant indicated in boldface. State clearly the mole ratio used for the conversion. a. \(2 \mathbf{M} g(s)+O_{2}(g) \rightarrow 2 M g O(s)\) b. \(2 \mathrm{Mg}(s)+\mathbf{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\) c. 4 Fe \((s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)\) d. \(4 \mathrm{Fe}(s)+3 \mathbf{O}_{2}(g) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\)

When the sugar glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is burned in air, carbon dioxide and water vapor are produced. Write the balanced chemical equation for this process, and calculate the theoretical yield of carbon dioxide when \(1.00 \mathrm{g}\) of glucose is burned completely.

The halogen elements are so reactive that the halides of many metals can be prepared by the direct combination of the elements. For example, iron(III) chloride can be prepared by the following reaction: $$2 \mathrm{Fe}(s)+3 \mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{FeCl}_{3}(s)$$ Calculate the mass of \(\mathrm{FeCl}_{3}\) that is formed if \(12.4 \mathrm{g}\) of iron reacts completely.
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