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

Hi everyone today, we have a question giving us the reaction copper sulfide solid plus two hydrochloric acid. Acquis forms copper chloride, acquis plus hydrogen sulfide gasses And it tells us that initially there are 1.34 moles of copper sulfide and 4.29 moles of hydrochloric acid. And we need to calculate the moles of the excess reactant that remains after this reaction is complete. So first we have to identify our limiting re agent and that is going to be copper sulfide because it has the least number of moles. Now we can put that into an equation. So 1.34 moles of copper sulfide times two moles of hydrochloric acid because that's how many we have in our balanced equation. Over one mole of copper sulfide Equals 2.68 moles of hydrochloric acid. So now we need to take our 4.29 moles of hydrochloric acid minus r 2.68 moles. And that gives us 1.61 moles of hydrochloric acid. So our answer is D 1.61 moles. Thank you for watching. Bye.

Ch.4 - Chemical Reactions and Chemical Quantities

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Tro 6th Edition

Ch.4 - Chemical Reactions and Chemical Quantities

Problem 49

Chapter 4, Problem 49

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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Identify the balanced chemical equation: 2 \text{ZnS}(s) + 3 \text{O}_2(g) \rightarrow 2 \text{ZnO}(s) + 2 \text{SO}_2(g).

Determine the mole ratio from the balanced equation: 2 moles of ZnS react with 3 moles of O_2.

Calculate the moles of O_2 needed to completely react with 4.2 moles of ZnS using the mole ratio: \frac{3 \text{ moles O}_2}{2 \text{ moles ZnS}}.

Calculate the moles of ZnS needed to completely react with 6.8 moles of O_2 using the mole ratio: \frac{2 \text{ moles ZnS}}{3 \text{ moles O}_2}.

Compare the calculated moles of O_2 and ZnS needed to determine the limiting reactant and calculate the remaining moles of the excess reactant.

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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 calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows us to determine the proportions of substances involved in a reaction, which is essential for identifying limiting and excess reactants. In this case, stoichiometry will help us find out how much of each reactant is consumed and how much remains after the reaction.

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 calculating the amounts of excess reactants left over after the reaction. In this scenario, we need to determine which reactant, ZnS or O2, limits the reaction to find the excess.

Excess Reactant

An excess reactant is a reactant that remains after the reaction has gone to completion because it was not completely consumed. Understanding the concept of excess reactants is important for calculating how much of a particular reactant is left over after the reaction. In this problem, we will calculate the moles of the excess reactant remaining after the reaction between ZnS and O2.

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Limiting Reagent Concept

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Textbook Question

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?

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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Textbook Question

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

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?

For the reaction shown, calculate the theoretical yield of product (in grams) for each initial amount of reactants. 2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s) c. 0.235 g Al, 1.15 g Cl₂

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Textbook Question

For the reaction shown, calculate the theoretical yield of the product (in grams) for each initial amount of reactants. Ti(s) + 2 F₂( g) → TiF₄(s) c. 0.233 g Ti, 0.288 g F₂

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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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Key Concepts

Stoichiometry

Limiting Reactant

Excess Reactant

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?