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 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 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

Hello, everyone. Today we have the following problem. Write the balanced chemical equation for each reaction. So we have B Cobalt three oxide reacts with hydrogen gas to form solid Cobalt and liquid water. So what does that look like? So we would have our Cobalt three oxide which is a solid reacting with hydrogen gas, which is of course, a gas and this will react to form solid Cobalt and liquid water. So we look at this equation and we see that it is not quite balanced. Of course, if we compare Adams on both sides, we can see that on the left, we have two Cobalts, but on the right, we have one on the left, we have three oxygens, but on the right, we have two, we have one and then hydrogens, we have two on the left and two on the right. So we need to balance the Cobalts to do that. We need to add a two to the coefficient of cobalt. And our equation, we multiply that one by the two to get two. And the next we balance our oxygens. We see that we have three on the left and one on the right. So we need to add a coefficient of three to our water. Such that when we multiply that oxygen by three, we have three, but we also change the number of hydrogens. So instead of two, we now have three times two or six. So what number can we multiply our hydrogen gas on the left to get to six, three. So if you multiply that by three, we get six and now it's balanced. And if we look at our anti choices, we see that choice c best reflects this and with that, we have solved the problem overall, I hope this helped. 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 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

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?

Textbook Question

For each of the reactions, calculate the mass (in grams) of the product that forms when 15.39 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.

a. 2 K(s) + Cl₂(g) → 2 KCl(s)

b. 2 K(s) + Br₂(l) → 2 KBr(s)

c. 4 Cr(s) + 3 O₂(g) → 2 Cr₂O₃(s)

d. 2 Sr(s) + O₂(g) → 2 SrO(s)

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Textbook Question

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?

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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