Consider the generic chemical equation: $$ 2 \mathrm{~A}+4 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ What is the limiting reactant when each of the initial quantities of \(\mathrm{A}\) and \(\mathrm{B}\) is allowed to react? (a) \(2 \mathrm{~mol} \mathrm{~A} ; 5 \mathrm{~mol} \mathrm{~B}\) (b) \(1.8 \mathrm{~mol} \mathrm{~A} ; 4 \mathrm{~mol} \mathrm{~B}\) (c) \(3 \mathrm{~mol} \mathrm{~A} ; 4 \mathrm{~mol} \mathrm{~B}\) (d) \(22 \mathrm{~mol} \mathrm{~A} ; 40 \mathrm{~mol} \mathrm{~B}\)

Short Answer

Expert verified
The limiting reactants are (a) A, (b) A, (c) B, (d) B.

Step by step solution

01

- Calculate the Mole Ratios

Use the coefficients in the chemical equation to determine the mole ratio of A to B. For the reaction, it's stated as 2 moles of A react with 4 moles of B. This ratio is 2:4, which simplifies to 1:2.
02

- Identify the Limiting Reactant for Scenario (a)

For every mole of A, you need 2 moles of B to fully react. For 2 moles of A provided, we need 4 moles of B. Given 2 moles A and 5 moles B, B is in excess since you would need only 4 moles of B to react with A. Thus, A is the limiting reactant.
03

- Identify the Limiting Reactant for Scenario (b)

Using the 1:2 ratio, 1.8 moles of A would need 3.6 moles of B to fully react. Given that there are 4 moles of B, B is again in excess, so A is the limiting reactant.
04

- Identify the Limiting Reactant for Scenario (c)

Using the 1:2 ratio, 3 moles of A would require 6 moles of B to completely react. Since only 4 moles of B are provided, B is not enough to react with all of A. Therefore, B is the limiting reactant.
05

- Identify the Limiting Reactant for Scenario (d)

With a 1:2 ratio, 22 moles of A would need 44 moles of B. Given 40 moles of B, there is not enough B to react with all of A, making B the limiting reactant.

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Key Concepts

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

Stoichiometry
Stoichiometry is the section of chemistry that involves quantitatively dealing with the relationships of the reactants and products in a chemical reaction. In simple terms, it's like a recipe that tells you how much of each ingredient you need to make a certain amount of a product. By using the coefficients from a balanced chemical equation, stoichiometry allows you to predict the amounts of substances consumed and produced in a reaction.

When solving a stoichiometry problem, the first step is to write a balanced chemical equation. This ensures that the number of atoms for each element is the same on both sides of the equation. Then, you identify the mole ratios from the equation, which compare the amounts of reactants and products in terms of moles—one of the central measurements in chemistry representing the amount of substance.

In the exercise provided, we use stoichiometry to determine the limiting reactant, the reactant that will be completely consumed first and thus will determine the amount of product formed. Recognizing the limiting reactant in a reaction is vital for calculations in chemistry to prevent waste of materials and to optimize the yield of the reaction.
Mole Ratio
The mole ratio is a critical concept derived from the coefficients of a balanced chemical equation. It provides the direct proportion of reactants that will react with each other, as well as the relationship between reactants and products. In our exercise, the mole ratio of substance A to substance B is determined from the chemical equation as 1:2, meaning that one mole of A reacts with two moles of B.

To apply the mole ratio, you simply use the ratio as a conversion factor to calculate the necessary amount of another reactant, or to predict the amount of product that will be formed when starting with a particular amount of reactant.

Importance of Understanding Mole Ratios

Without an understanding of mole ratios, it would be difficult to scale up reactions for commercial purposes or to know how much of each reactant is required for a laboratory experiment. For students learning to identify the limiting reactant, understanding mole ratios ensures accurate and relevant chemical calculations.
Chemical Equations
Chemical equations are symbolic representations of chemical reactions. They show the reactants (the starting substances) turning into the products (resulting substances) through a chemical process. A balanced chemical equation has equal numbers of each type of atom on both sides, thereby satisfying the law of conservation of mass.

The coefficients in a chemical equation, such as the '2' in front of substance A and the '4' in front of substance B in our example, indicate the mole ratios in which the substances react. These coefficients are indispensable in performing stoichiometric calculations and in identifying the limiting reactant.

Real-world Application of Chemical Equations

Understanding chemical equations not only helps in academic settings but is also crucial for industrial applications, where the precise combination of reactants can have economic and safety implications. Additionally, in research, correct interpretation of chemical equations can lead to the development of new materials and medicines.

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Most popular questions from this chapter

Consider the combustion of propane: $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Balance the reaction. (b) Divide all coefficients by the coefficient on propane, so that you have the reaction for the combustion of 1 mol of propane. (c) \(\Delta H_{\mathrm{rxn}}\) for the combustion of one mole of propane is \(-2219 \mathrm{~kJ}\). What mass of propane would you need to burn to generate \(5.0 \mathrm{MJ}\) of heat? (d) If propane costs about \(\$ 0.67 / \mathrm{L}\) and has a density of \(2.01 \mathrm{~g} / \mathrm{cm}^{3}\), how much would it cost to generate \(5.0 \mathrm{MJ}\) of heat by burning propane?

The combustion of liquid ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) produces carbon dioxide and water. After \(3.8 \mathrm{~mL}\) of ethanol (density \(=0.789 \mathrm{~g} / \mathrm{mL}\) ) is allowed to burn in the presence of \(12.5 \mathrm{~g}\) of oxygen gas, \(3.10 \mathrm{~mL}\) of water (density \(=1.00 \mathrm{~g} / \mathrm{mL}\) ) is collected. Determine the limiting reactant, theoretical yield of \(\mathrm{H}_{2} \mathrm{O}\), and percent yield for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)

Consider the generic reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C} \Delta H_{\mathrm{rxn}}=-125 \mathrm{~kJ} $$ Determine the amount of heat emitted when each amount of reactant completely reacts (assume that there is more than enough of the other reactant). (a) \(2 \mathrm{~mol} \mathrm{~A}\) (b) \(3 \mathrm{~mol} \mathrm{~A}\) (c) \(3 \mathrm{~mol} \mathrm{~B}\) (d) \(5 \mathrm{~mol} \mathrm{~B}\)

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. $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) \(4.6 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(4.6 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(0.0558 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\) (d) \(0.0558 \mathrm{~mol} \mathrm{O}_{2}\)

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) \(4 \mathrm{~mol} \mathrm{~A} ; 5 \mathrm{~mol} \mathrm{~B}\)

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