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Chapter 3: Problem 29

What is the theoretical yield for a reaction, and how does this quantity depend on the limiting reactant?

Short Answer

Expert verified
The theoretical yield in a reaction is the maximum amount of product that can be formed, based on the balanced chemical equation and initial amounts of reactants. It depends on the limiting reactant, which is the reactant that is completely consumed during the reaction and determines the maximum amount of product that can be formed. In the example of the reaction \(2H_2 + O_2 \rightarrow 2H_2O\), where we have 4 moles of H2 and 2 moles of O2, both reactants are limiting, and the theoretical yield of water is 4 moles.

Step by step solution

01

1. Defining Theoretical Yield

Theoretical yield is the maximum amount of product that can be formed in a chemical reaction, based on the balanced chemical equation and the initial amounts of the reactants. In any reaction, the theoretical yield is calculated by considering the stoichiometric ratio or mole-to-mole ratio between the reactants and products.
02

2. Understanding Limiting Reactant

The limiting reactant is the reactant that is completely consumed during a chemical reaction and determines the maximum amount of product that can be formed. It limits the reaction progress because once it is fully consumed, the reaction cannot proceed further, even if there are still other reactants available.
03

3. Choosing a Sample Reaction

Let's use the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O) as an example: \[ 2H_2 + O_2 \rightarrow 2H_2O \]
04

4. Calculating Mole-to-Mole Ratio

From the balanced chemical equation, we can see that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. This gives us a mole-to-mole ratio of 2:1:2 for H2:O2:H2O.
05

5. Identifying Limiting Reactant

Suppose we have 4 moles of H2 and 2 moles of O2, we need to determine the limiting reactant by comparing the relative amounts of each reactant to the mole-to-mole ratio. Divide the moles of each reactant by their respective coefficients in the balanced equation: - For H2: 4 moles H2 / 2 = 2 - For O2: 2 moles O2 / 1 = 2 Both values are equal, indicating that both reactants are present in the correct stoichiometric ratio and both are limiting reactants.
06

6. Calculating Theoretical Yield

To calculate the theoretical yield, use the limiting reactant value and the mole-to-mole ratio between the limiting reactant and the product. In this case, since both reactants are limiting, either can be used. We'll choose H2 for our calculation: - Theoretical Yield (H2O) = 2 (moles of H2) 脳 (2 moles of H2O / 2 moles of H2) - Theoretical Yield (H2O) = 4 moles of H2O The theoretical yield of water in this reaction is 4 moles. This yield depends on the limiting reactant, which in this case was determined to be both H2 and O2.

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Most popular questions from this chapter

Methane \(\left(\mathrm{CH}_{4}\right)\) is the main component of marsh gas. Heating methane in the presence of sulfur produces carbon disulfide and hydrogen sulfide as the only products. a. Write the balanced chemical equation for the reaction of methane and sulfur. b. Calculate the theoretical yield of carbon disulfide when \(120 . \mathrm{g}\) of methane is reacted with an equal mass of sulfur.

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g}\) \(\mathrm{CO}\), and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Chloral hydrate \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\right)\) is a drug formerly used as a sedative and hypnotic. It is the compound used to make "Mickey Finns" in detective stories. a. Calculate the molar mass of chloral hydrate. b. What amount (moles) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\) molecules are in \(500.0 \mathrm{~g}\) chloral hydrate? c. What is the mass in grams of \(2.0 \times 10^{-2}\) mol chloral hydrate? d. What number of chlorine atoms are in \(5.0 \mathrm{~g}\) chloral hydrate? e. What mass of chloral hydrate would contain \(1.0 \mathrm{~g} \mathrm{Cl} ?\) \(\mathbf{f}\). What is the mass of exactly 500 molecules of chloral hydrate?
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