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Chapter 6: Problem 45

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 \(10.0 \mathrm{~g} \mathrm{Fe}\) and \(2.00 \mathrm{~g} \mathrm{O}_{2}\) are reacted?

Short Answer

Expert verified
The short answers for each part of the question are: a. -1652 kJ of heat is released when 4.00 moles of iron are reacted with excess O₂. b. -826 kJ of heat is released when 1.00 mole of Fe₂O₃ is produced. c. -7.38 kJ of heat is released when 1.00 g iron is reacted with excess O₂. d. -34.4 kJ of heat is released when 10.0 g Fe and 2.00 g O₂ are reacted.

Step by step solution

01

We are given 4.00 moles of iron, with excess O₂. The enthalpy change for the reaction, ΔH, is -1652 kJ. #Step 2: Use stoichiometry to determine the amount of heat released#

According to the balanced equation, 4 moles of Fe react with 3 moles of O₂ to produce 2 moles of Fe₂O₃ and release -1652 kJ of heat. Since the moles of iron given in the problem are equal to the moles in the balanced equation, the heat released remains the same as the ΔH value given. Therefore, the heat released is -1652 kJ. b. How much heat is released when 1.00 mole of Fe₂O₃ is produced? #Step 1: Identify the given information#
02

We are given the production of 1.00 mole of Fe₂O₃. The enthalpy change for the reaction, ΔH, is -1652 kJ. #Step 2: Use stoichiometry to determine the amount of heat released#

According to the balanced equation, 2 moles of Fe₂O₃ are produced with -1652 kJ heat release. To determine the heat released when only 1 mole is produced, divide ΔH by 2: Heat released = (-1652 kJ) / 2 = -826 kJ c. How much heat is released when 1.00 g iron is reacted with excess O₂? #Step 1: Convert grams of iron to moles of iron#
03

Use the molar mass of iron (Fe) to convert grams to moles: Moles of Fe = (1.00 g) / (55.85 g/mol) ≈ 0.0179 mol #Step 2: Use stoichiometry to determine the amount of heat released per mole of iron#

Since 4 moles of iron react to release -1652 kJ of heat, determine the heat per mole of iron: Heat per mole of iron = (-1652 kJ) / 4 = -413 kJ/mol #Step 3: Calculate the heat released for 1.00 g of iron#
04

Multiply the moles of iron by the heat released per mole of iron. Heat released = 0.0179 mol × -413 kJ/mol ≈ -7.38 kJ d. How much heat is released when 10.0 g Fe and 2.00 g O₂ are reacted? #Step 1: Determine the limiting reactant#

Convert the given mass of each reactant to moles using their respective molar masses (Fe: 55.85 g/mol, O₂: 32.00 g/mol): Moles of Fe = (10.0 g) / (55.85 g/mol) ≈ 0.179 mol Moles of O₂ = (2.00 g) / (32.00 g/mol) ≈ 0.0625 mol In the balanced equation, the stoichiometric ratio between Fe and O₂ is 4:3. Divide the moles of each reactant by the stoichiometric coefficients: Fe: 0.179 / 4 ≈ 0.0447 O₂: 0.0625 / 3 ≈ 0.0208 Since the quotient for O₂ is smaller than the quotient for Fe, O₂ is the limiting reactant. #Step 2: Calculate the heat released based on the limiting reactant#
05

Based on the enthalpy change, -1652 kJ is released for every 3 moles of O₂ reacted. Find the heat released per mole of O₂, so divide ΔH by 3: Heat per mole of O₂ = (-1652 kJ) / 3 = -550.67 kJ/mol #Step 3: Calculate the heat released for the given amount of O₂#

Multiply the moles of O₂ by the heat released per mole of O₂: Heat released = 0.0625 mol × -550.67 kJ/mol ≈ -34.4 kJ

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