Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O 2 ( g) → 2 Al 2 O 3 (s) a. 4 mol Al, 2 mol O 2 b. 5 mol Al, 3 mol O 2 c. 17 mol Al, 11 mol O 2 d. 8.8 mol Al, 7.2 mol O 2

1. The balanced chemical equation is 4 Al(s) + 3 O_{2}( g) → 2 $$Al_{2}O$$_{3}(s). This tells us that for every 4 moles of Al, we need 3 moles of O_{2} to completely react.. 2. For each scenario, calculate the mole ratio of Al to O_{2} and compare it to the stoichiometric ratio from the balanced equation. The reactant that is less than or equal to its stoichiometric ratio is the limiting reactant.. 3. For scenario a, the mole ratio of Al to O_{2} is 4/2 = 2. The stoichiometric ratio from the balanced equation is 4/3. Since 2 \u003c 4/3, O_{2} is the limiting reactant.. 4. Repeat step 3 for scenarios b, c, and d. For scenario b, the mole ratio of Al to O_{2} is 5/3. For scenario c, the mole ratio of Al to O_{2} is 17/11. For scenario d, the mole ratio of Al to O_{2} is 8.8/7.2.. 5. In each case, the reactant that has a mole ratio less than or equal to the stoichiometric ratio from the balanced equation is the limiting reactant. This is the reactant that will be completely consumed first in the reaction.

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O2( g) → 2 Al2O3(s) a. 4 mol Al, 2 mol O2 b. 5 mol Al, 3 mol O2 c. 17 mol Al, 11 mol O2 d. 8.8 mol Al, 7.2 mol O2

1. The balanced chemical equation is 4 Al(s) + 3 O_{2}( g) → 2 $$Al_{2}O$$_{3}(s). This tells us that for every 4 moles of Al, we need 3 moles of O_{2} to completely react. 2. For each scenario, calculate the mole ratio of Al to O_{2} and compare it to the stoichiometric ratio from the balanced equation. The reactant that is less than or equal to its stoichiometric ratio is the limiting reactant. 3. For scenario a, the mole ratio of Al to O_{2} is 4/2 = 2. The stoichiometric ratio from the balanced equation is 4/3. Since 2 \u003c 4/3, O_{2} is the limiting reactant. 4. Repeat step 3 for scenarios b, c, and d. For scenario b, the mole ratio of Al to O_{2} is 5/3. For scenario c, the mole ratio of Al to O_{2} is 17/11. For scenario d, the mole ratio of Al to O_{2} is 8.8/7.2. 5. In each case, the reactant that has a mole ratio less than or equal to the stoichiometric ratio from the balanced equation is the limiting reactant. This is the reactant that will be completely consumed first in the reaction.

Ch.4 - Chemical Reactions and Chemical Quantities

Tro - Chemistry: A Molecular Approach 6th Edition

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Tro 6th Edition

Ch.4 - Chemical Reactions and Chemical Quantities

Problem 44

Chapter 4, Problem 44

Find the limiting reactant for each initial amount of reactants. 4 Al(s) + 3 O₂( g) → 2 Al₂O₃(s)
a. 4 mol Al, 2 mol O₂
b. 5 mol Al, 3 mol O₂
c. 17 mol Al, 11 mol O₂
d. 8.8 mol Al, 7.2 mol O₂

Verified step by step guidance

1. The balanced chemical equation is 4 Al(s) + 3 O₂( g) → 2 Al₂O₃(s). This tells us that for every 4 moles of Al, we need 3 moles of O₂ to completely react.

2. For each scenario, calculate the mole ratio of Al to O₂ and compare it to the stoichiometric ratio from the balanced equation. The reactant that is less than or equal to its stoichiometric ratio is the limiting reactant.

3. For scenario a, the mole ratio of Al to O₂ is 4/2 = 2. The stoichiometric ratio from the balanced equation is 4/3. Since 2 < 4/3, O₂ is the limiting reactant.

4. Repeat step 3 for scenarios b, c, and d. For scenario b, the mole ratio of Al to O₂ is 5/3. For scenario c, the mole ratio of Al to O₂ is 17/11. For scenario d, the mole ratio of Al to O₂ is 8.8/7.2.

5. In each case, the reactant that has a mole ratio less than or equal to the stoichiometric ratio from the balanced equation is the limiting reactant. This is the reactant that will be completely consumed first in the reaction.

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Key Concepts

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

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. To identify it, one must compare the mole ratios of the reactants used in the reaction with the coefficients from the balanced chemical equation.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. It involves using mole ratios derived from the coefficients of the balanced equation to determine how much of each reactant is needed or how much product can be produced.

Mole Ratio

Mole ratio is the ratio of moles of one substance to moles of another substance in a balanced chemical equation. It is essential for converting between moles of reactants and products, allowing chemists to predict how much of each reactant is required to fully react or how much product will be formed.

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Textbook Question

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

Find the limiting reactant for each initial amount of reactants. 2 Na(s) + Br₂( g) → 2 NaBr(s) c. 1.5 mol Na, 2.1 mol Br₂

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Textbook Question

Consider the reaction: 4 HCl(g) + O₂(g) → 2 H₂O(g) + 2 Cl₂(g) Each molecular diagram represents an initial mixture of reactants. How many molecules of Cl₂ form from the reaction mixture that produces the greatest amount of products?

Sulfuric acid dissolves aluminum metal according to the reaction:

2 Al(s) + 3 H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3 H₂(g)

Suppose you want to dissolve an aluminum block with a mass of 11.3 g. What minimum mass of H₂SO₄ (in g) do you need? What mass of H₂ gas (in g) does the complete reaction of the aluminum block produce?