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The overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{kJ} $$ a. How much heat is released when 4.00 moles of iron are reacted with excess \(\mathrm{O}_{2} ?\) b. How much heat is released when 1.00 mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when 1.00 \(\mathrm{g}\) iron is reacted with excess \(\mathrm{O}_{2} ?\) d. How much heat is released when 10.0 \(\mathrm{g}\) Fe and 2.00 \(\mathrm{g} \mathrm{O}_{2}\) are reacted?

Short Answer

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a. Since 4 moles of Fe release -1652 kJ, the heat released when 4.00 moles of iron are reacted with excess O₂ is -1652 kJ. b. Since 2 moles of Fe₂O₃ are produced when -1652 kJ of heat is released, the heat released when 1.00 mole of Fe₂O₃ is produced is -826 kJ. c. For 1.00 g of iron, the heat released when reacted with excess O₂ is -7.52 kJ. d. When 10.0 g Fe and 2.00 g O₂ are reacted, the heat released is -34.3 kJ.

Step by step solution

01

Observe the stoichiometry of the reaction

The balanced chemical equation shows that 4 moles of Fe react to release -1652 kJ of heat.
02

Calculate the heat released

By observing the stoichiometry, the heat released when 4 moles of Fe react can be calculated as: \[ \text{Heat released} = -1652 \, \text{kJ} \] Since the same amount of moles of Fe is provided as per the balanced equation, then the heat released is -1652 kJ. b. How much heat is released when 1.00 mole of Fe₂O₃ is produced?
03

Observe the stoichiometry of the reaction

The balanced chemical equation shows that 2 moles of Fe₂O₃ are produced when the reaction releases -1652 kJ of heat.
04

Calculate the heat released

Based on the stoichiometry, we can find the heat released when 1 mole of Fe₂O₃ is formed as: \[ \text{Heat released} = -\frac{1652}{2} \, \text{kJ} = -826 \, \text{kJ} \] c. How much heat is released when 1.00 g iron is reacted with excess O₂?
05

Calculate the moles of Fe

First, determine the number of moles of Fe in 1.00 g: \[ \text{Moles of Fe} = \frac{\text{mass of Fe}}{\text{molar mass of Fe}} = \frac{1.00 \, \text{g}}{55.85 \, \text{g/mol}} = 0.0179 \, \text{mol} \]
06

Calculate the heat released

In the next step, we can find the heat released when 0.0179 moles of Fe react by using the stoichiometry ratio and the heat released by 4 moles of Fe: \[ \text{Heat released} = \frac{0.0179 \, \text{mol}}{4.00 \, \text{mol}} \times -1652 \, \text{kJ} = -7.52 \, \text{kJ} \] d. How much heat is released when 10.0 g Fe and 2.00 g Oâ‚‚ are reacted?
07

Calculate the moles of Fe and Oâ‚‚

Calculate the number of moles of Fe and Oâ‚‚ involved: \[ \text{Moles of Fe} = \frac{10.0 \, \text{g}}{55.85 \, \text{g/mol}} = 0.179 \, \text{mol} \] \[ \text{Moles of Oâ‚‚} = \frac{2.00 \, \text{g}}{32.00 \, \text{g/mol}} = 0.0625 \, \text{mol} \]
08

Identify the limiting reactant

Determine the limiting reactant by comparing the moles of Fe and Oâ‚‚ with their stoichiometric coefficients (4 for Fe and 3 for Oâ‚‚): \[ \frac{\text{moles of Fe}}{4} = \frac{0.179 \, \text{mol}}{4} = 0.0448 \, \text{mol} \] \[ \frac{\text{moles of Oâ‚‚}}{3} = \frac{0.0625 \, \text{mol}}{3} = 0.0208 \, \text{mol} \] Since 0.0208 < 0.0448, Oâ‚‚ is the limiting reactant.
09

Calculate the heat released

Now we can determine the heat released when 0.0625 moles of Oâ‚‚ react with excess Fe, using the stoichiometry ratio and the heat released by 3 moles of Oâ‚‚: \[ \text{Heat released} = \frac{0.0625 \, \text{mol}}{3.00 \, \text{mol}} \times -1652 \, \text{kJ} = -34.3 \, \text{kJ} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermochemistry: Understanding Enthalpy in Reactions
Thermochemistry is a branch of chemistry focused on the study of heat energy associated with chemical reactions and physical transformations. Every chemical reaction involves a change in enthalpy, which is the total heat content of the system. This is denoted as \( \Delta H \) and can be either positive or negative, depending on the nature of the reaction.
In exothermic reactions, like the one in the heat pack using iron and oxygen, energy is released, resulting in a negative \( \Delta H \). The given reaction releases \(-1652 \text{ kJ}\) when 4 moles of Fe react with 3 moles of \( \text{O}_{2} \). The energy released is due to the formation of stronger bonds in the products compared to the reactants.
Understanding the enthalpy change of a reaction helps predict whether energy will be absorbed or released, which is crucial in applications such as heat packs where heat release is desired.
Stoichiometry: Balancing Chemical Equations
Stoichiometry is the calculation of reactants and products in chemical reactions. It ensures that a chemical equation is balanced, meaning the same number of each type of atom appears on both sides of the equation. This balance reflects the law of conservation of mass: matter is neither created nor destroyed.
In the exercise, the balanced equation \( 4 \text{Fe} + 3 \text{O}_{2} \rightarrow 2 \text{Fe}_{2}\text{O}_{3} \) highlights that for every 4 moles of Fe, 3 moles of \( \text{O}_{2} \) are required to produce 2 moles of \( \text{Fe}_{2}\text{O}_{3} \). Stoichiometry allows us to calculate heat produced per mole, such as \(-1652\, \text{kJ}\) for 4 moles Fe, or \(-826\, \text{kJ} \) for 1 mole of \( \text{Fe}_{2}\text{O}_{3} \).
Accurate stoichiometric calculations are essential for predicting the outcome of reactions, managing reactants, and ensuring conditions yielding desired quantities of products.
Limiting Reactant: Determining Reaction Output
The limiting reactant in a chemical equation is the substance that is entirely consumed first, stopping the reaction from continuing. Identifying the limiting reactant is crucial, especially in calculations involving heat transfer, because it dictates the maximum amount of product formed and energy change.
From the exercise, when 10.0 g Fe and 2.00 g \( \text{O}_{2} \) are used, we calculate moles to find that \( \text{O}_{2} \) is the limiting reactant, dictating the reaction's extent. Using stoichiometry, dividing the moles of each reactant by its coefficient in the balanced equation reveals the actual limiting substance.
This determination guides further calculations like how much heat is released, as any additional reactant does not change the outcome unless more of the limiting reactant is added.
Heat Transfer: Quantifying Energy Exchange
Heat transfer in chemical reactions refers to the movement of thermal energy from one substance to another, or to its surroundings. In the context of thermochemical equations, knowing how much heat is released or absorbed is fundamental.
For example, reacting a specific mass of a substance, like 1.00 g of iron, involves converting the mass to moles, and using stoichiometry to find associated heat transfer. In this case, \( 0.0179 \) moles of Fe yield \( -7.52 \text{ kJ} \) of heat. This systematic approach can be applied regardless of reactant quantity, consistently calculating the thermal energy exchange.
  • Converting grams to moles helps understand how smaller quantities affect heat release.
  • Using molar ratios calculates how much heat energy each mole produces or consumes in the overall change.
Knowing these heat transfer details is essential in designing systems where energy management is critical, such as heating elements or thermal stabilizers.

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Most popular questions from this chapter

Calculate the internal energy change for each of the following. a. One hundred \((100 .)\) joules of work is required to compress a gas. At the same time, the gas releases 23 \(\mathrm{J}\) of heat. b. A piston is compressed from a volume of 8.30 \(\mathrm{L}\) to 2.80 \(\mathrm{L}\) against a constant pressure of 1.90 \(\mathrm{atm}\) . In the process, there is a heat gain by the system of 350. J. c. A piston expands against 1.00 atm of pressure from 11.2 \(\mathrm{L}\) to 29.1 \(\mathrm{L}\) . In the process, 1037 \(\mathrm{J}\) of heat is absorbed.

Consider the following reaction: $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-572 \mathrm{kJ} $$ a. How much heat is evolved for the production of 1.00 mole of \(\mathrm{H}_{2} \mathrm{O}(l) ?\) b. How much heat is evolved when 4.03 g hydrogen are reacted with excess oxygen? c. How much heat is evolved when 186 \(\mathrm{g}\) oxygen are reacted with excess hydrogen? d. The total volume of hydrogen gas needed to fill the Hindenburg was \(2.0 \times 10^{8} \mathrm{L}\) at 1.0 atm and \(25^{\circ} \mathrm{C} .\) How much heat was evolved when the Hindenburg exploded, assuming all of the hydrogen reacted?

Standard enthalpies of formation are relative values. What are \(\Delta H_{\mathrm{f}}^{\circ}\) values relative to?

The reaction $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is \(-227 \mathrm{kJ} .\) In designing a sulfuric acid plant, is it necessary to provide for heating or cooling of the reaction mixture? Explain.

Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for \(\mathrm{C}_{4} \mathrm{H}_{4}(-2341 \mathrm{kJ} / \mathrm{mol}), \mathrm{C}_{4} \mathrm{H}_{8}\) \((-2755 \mathrm{kJ} / \mathrm{mol}),\) and \(\mathrm{H}_{2}(-286 \mathrm{kJ} / \mathrm{mol}),\) calculate \(\Delta H\) for the reaction $$ \mathrm{C}_{4} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{8}(g) $$

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