Determine the theoretical yield of \(\mathrm{C}\) when each of the initial quantities of \(\mathrm{A}\) and \(\mathrm{B}\) is allowed to react in the generic reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 2 \mathrm{C} $$ (a) \(2 \mathrm{~mol} \mathrm{~A} ; 4 \mathrm{~mol} \mathrm{~B}\) (b) \(3 \mathrm{~mol} \mathrm{~A} ; 3 \mathrm{~mol} \mathrm{~B}\) (c) \(5 \mathrm{~mol} \mathrm{~A} ; 6 \mathrm{~mol} \mathrm{~B}\) (d) \(4 \mathrm{~mol} \mathrm{~A} ; 5 \mathrm{~mol} \mathrm{~B}\)

Short Answer

Expert verified
The theoretical yields are (a) 2 mol C, (b) 2 mol C, (c) 4 mol C, (d) 3.33 mol C.

Step by step solution

01

Identify the Limiting Reactant for each case

To find the theoretical yield of C, determine which reactant (A or B) is the limiting reactant for each case. This is done by dividing the moles of each reactant by their respective coefficients in the balanced equation and finding the smallest quotient for each case.
02

Calculate the Theoretical Yield of C

Once the limiting reactant is identified, use stoichiometry to calculate the moles of C produced based on the limiting reactant. According to the reaction stoichiometry, 2 moles of A or 3 moles of B produce 2 moles of C. Use the number of moles of the limiting reactant to find the moles of C produced.
03

Repeat for Each Case (a-d)

Perform the above steps for each case (a-d) to find the theoretical yields of C.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limiting Reactant Determination
Understanding which reactant limits a chemical reaction is crucial in predicting how much product can be formed. The limiting reactant is the substance that is completely consumed first during the chemical reaction, causing the reaction to stop and determining the maximum amount of product that can be formed.

To determine the limiting reactant, we compare the mole ratios of reactants to the coefficients in the balanced chemical equation. We calculate the molar ratio of each reactant to its coefficient and the smallest of these ratios indicates the limiting reactant. For instance, in the given reaction \(2A + 3B \rightarrow 2C\), we divide the moles of A and B by their respective coefficients. Thus, for case (a), we do \(\frac{2 \text{ mol A}}{2}\) and \(\frac{4 \text{ mol B}}{3}\) to find which is smaller. By consistently applying this method, we can identify which reactant will be exhausted first, and consequently, control the output of the reaction.
Stoichiometry
Stoichiometry is the method of quantitatively analyzing the proportions of elements and compounds involved in chemical reactions. It is based on the conservation of mass and the law of definite proportions, which state that in a chemical reaction, elements are combined in fixed ratios by mass, equal to the ratios of their mole amounts.

Stoichiometry allows us to predict the results of chemical reactions, including how much product will form from given amounts of reactants. Calculating stoichiometric quantities requires a balanced chemical equation, as it provides the mole ratios we need to convert between reactants and products. For example, according to the balanced reaction in our exercise, 2 moles of A react with 3 moles of B to produce 2 moles of C. This fixed ratio means that the amount of product C can be directly calculated from the amount of the limiting reactant using stoichiometry.
Chemical Reaction Balancing
Balancing a chemical reaction is a fundamental skill in chemistry as it ensures that the law of conservation of mass is respected. It means adjusting the coefficients of the reactants and products so that the number of atoms of each element is the same on both sides of the reaction equation.

Using the generic balanced reaction from our exercise \(2A + 3B \rightarrow 2C\), we can see that for every 2 moles of A and 3 moles of B reacted, 2 moles of C are produced. Balancing chemical equations is vital because it lays the groundwork for stoichiometric calculations, allowing us to accurately determine the theoretical yields of products in a chemical reaction. In studying reaction balancing, one also learns to systematically approach problems and apply mathematical reasoning to the physical world.
Mole-to-Mole Ratio
The mole-to-mole ratio, derived from the coefficients of a balanced chemical equation, serves as a conversion factor between the amount of reactants and products. It's an essential concept in stoichiometry as it quantifies the direct relationship between the substances in a chemical reaction.

Continuing with our reaction \(2A + 3B \rightarrow 2C\), the coefficients indicate the mole-to-mole ratios: for every 2 moles of A, we can produce 2 moles of C, and for every 3 moles of B, 2 moles of C can be produced. These ratios allow us to calculate the amount of products formed from a given amount of reactants. Always remember, it is the limiting reactant's mole-to-mole ratio to the product that informs the maximum theoretical yield of the product in any given chemical reaction.

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Most popular questions from this chapter

Determine the theoretical yield of \(C\) when each of the initial quantities of \(\mathrm{A}\) and \(\mathrm{B}\) is allowed to react in the generic reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ (a) \(1 \mathrm{~mol} \mathrm{~A} ; 1 \mathrm{~mol} \mathrm{~B}\) (b) \(2 \mathrm{~mol} \mathrm{~A} ; 2 \mathrm{~mol} \mathrm{~B}\) (c) \(1 \mathrm{~mol} \mathrm{~A} ; 3 \mathrm{~mol} \mathrm{~B}\) (d) \(32 \mathrm{~mol} \mathrm{~A} ; 68 \mathrm{~mol} \mathrm{~B}\)

Hydrochloric acid can dissolve solid iron according to the reaction: $$ \mathrm{Fe}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{FeCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ What minimum mass of \(\mathrm{HCl}\) in grams dissolves a \(2.8\)-g iron bar on a padlock? How much \(\mathrm{H}_{2}\) is produced by the complete reaction of the iron bar?

For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume there is more than enough of the other reactant. $$ 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) $$ (a) \(2.4 \mathrm{~mol} \mathrm{PbS}\) (b) \(2.4 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(5.3 \mathrm{~mol} \mathrm{PbS}\) (d) \(5.3 \mathrm{~mol} \mathrm{O}_{2}\)

Write the conversion factor that you would use to convert from moles of \(\mathrm{Cl}_{2}\) to moles of \(\mathrm{NaCl}\) in the reaction: $$ 2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s) $$

Consider the reaction between reactants \(\mathrm{S}\) and \(\mathrm{O}_{2}\) : $$ 2 \mathrm{~S}(\mathrm{~s})+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) $$ If a reaction vessel initially contains \(5 \mathrm{~mol} \mathrm{~S}\) and \(9 \mathrm{~mol} \mathrm{O}_{2}\), how many moles of \(\mathrm{S}, \mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\) will be in the reaction vessel after the reactants have reacted as much as possible? (Assume \(100 \%\) actual yield.)

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