Calculate the theoretical yield of product (in moles) for each initial amount of reactants. 3 Mn(s) + 2 O 2 (g) → Mn 3 O 4 (s) a. 3 mol Mn, 2 mol O 2 b. 4 mol Mn, 7 mol O 2 c. 27.5 mol Mn, 43.8 mol O 2

Hello, everyone. Today, we have the following problem, calculate the theoretically yield of the product in moles for each initial amount of reactants. So we have to use the mole ratios to convert the reactants into the product. And so we are given our four moles of our manganese and seven moles of oxygen. So for our manganese, we take our four moles of it and we have to multiply by the multiple ratio. That is saying that if we look at our equation, we see how many moles of manganese we have, which is three, we put that in the denominator so that we can cancel it and it's out. And then on our enumerator, we put the moles of our manganese oxide that we formed, which is just one. And we see now when our units cancel L we write an answer of 1.333 moles of that manganese oxide. We do the same for our oxygen. So we have our seven moles of our oxygen gas. And we do the same thing with our multiple conversion. We say that we have one mole of our manganese sign in ratio with our two moles of oxygen gas, our units will cancel out and this will equal 3.5 moles of our manganese oxide. Since manganese produce the less or not, it is the limiting reactant. Well, oxygen gas of the excess and the product produced when the limiting reactant fully reacted is the theoretical yield. And so essentially, when we use this 1.33 moles of our manganese oxide, we yielded one mole of our manganese oxide or our product making your answer and your choice. A and with that, we have solved the problem overall, I hope is help. And until next time.

Ch.4 - Chemical Reactions and Chemical Quantities

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Tro 6th Edition

Ch.4 - Chemical Reactions and Chemical Quantities

Problem 48

Chapter 4, Problem 48

Calculate the theoretical yield of product (in moles) for each initial amount of reactants.
3 Mn(s) + 2 O₂(g) → Mn₃O₄(s)
a. 3 mol Mn, 2 mol O₂
b. 4 mol Mn, 7 mol O₂
c. 27.5 mol Mn, 43.8 mol O₂

Verified step by step guidance

Identify the balanced chemical equation: 3 \text{ Mn}(s) + 2 \text{ O}_2(g) \rightarrow \text{ Mn}_3\text{O}_4(s)

Determine the molar ratio between reactants and product from the balanced equation: 3 \text{ mol Mn} : 2 \text{ mol O}_2 : 1 \text{ mol Mn}_3\text{O}_4

Calculate the theoretical yield of Mn_3O_4 based on the amount of Mn: \frac{4 \text{ mol Mn}}{3 \text{ mol Mn}} \times 1 \text{ mol Mn}_3\text{O}_4

Calculate the theoretical yield of Mn_3O_4 based on the amount of O_2: \frac{7 \text{ mol O}_2}{2 \text{ mol O}_2} \times 1 \text{ mol Mn}_3\text{O}_4

Identify the limiting reactant by comparing the theoretical yields from Mn and O_2, and use the smaller yield as the theoretical yield of Mn_3O_4.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction based on balanced chemical equations. Understanding stoichiometry is essential for calculating theoretical yields, as it provides the mole ratios needed to determine how much product can be formed from specific amounts of reactants.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. Identifying the limiting reactant is crucial for yield calculations, as it dictates the extent of the reaction. In the given question, one must compare the mole ratios of the reactants to find which one limits the formation of Mn3O4.

Theoretical Yield

Theoretical yield is the maximum amount of product that can be generated from a given amount of reactants, assuming complete conversion and no losses. It is calculated using stoichiometric relationships derived from the balanced chemical equation. The theoretical yield is important for evaluating the efficiency of a reaction and is often compared to the actual yield obtained in practice to determine the reaction's success.

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Consider the reaction: 2 CH₃OH(g) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g) Each of the molecular diagrams represents an initial mixture of the reactants. How many CO₂ molecules form from the reaction mixture that produces the greatest amount of products?

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Calculate the theoretical yield of the product (in moles) for each initial amount of reactants.

Ti(s) + 2 Cl₂(g) → TiCl₄(s)

a. 4 mol Ti, 4 mol Cl₂

b. 7 mol Ti, 17 mol Cl₂

c. 12.4 mol Ti, 18.8 mol Cl₂

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Zinc sulfide reacts with oxygen according to the reaction: 2 ZnS(s) + 3 O₂(g) → 2 ZnO(s) + 2 SO₂( g) A reaction mixture initially contains 4.2 mol ZnS and 6.8 mol O₂. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

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Iron(II) sulfide reacts with hydrochloric acid according to the reaction: FeS(s) + 2 HCl(aq) → FeCl₂(s) + H₂S(g) A reaction mixture initially contains 0.223 mol FeS and 0.652 mol HCl. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant remains?

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Calculate the theoretical yield of product (in moles) for each initial amount of reactants. 3 Mn(s) + 2 O2(g) → Mn3O4(s) a. 3 mol Mn, 2 mol O2b. 4 mol Mn, 7 mol O2c. 27.5 mol Mn, 43.8 mol O2

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Limiting Reactant

Theoretical Yield

Calculate the theoretical yield of product (in moles) for each initial amount of reactants.
3 Mn(s) + 2 O₂(g) → Mn₃O₄(s)
a. 3 mol Mn, 2 mol O₂
b. 4 mol Mn, 7 mol O₂
c. 27.5 mol Mn, 43.8 mol O₂