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Chapter 7: 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 \(\mathbf{O}_{2} ?\) d. How much heat is released when \(10.0 \mathrm{g}\) Fe and \(2.00 \mathrm{g} \mathrm{O}_{2}\) are reacted?

Short Answer

Expert verified
a. Heat released when 4.00 moles of iron are reacted with excess O₂: -1652 kJ. b. Heat released when 1.00 mole of Fe₂O₃ is produced: -826 kJ. c. Heat released when 1.00 g iron is reacted with excess O₂: -7.49 kJ. d. Heat released when 10.0 g Fe and 2.00 g O₂ are reacted: -34.39 kJ.

Step by step solution

01

Write down the balanced equation and the enthalpy change

The balanced equation: \(4 \textrm{Fe}(s) + 3 \textrm{O}_{2}(g) \rightarrow 2 \textrm{Fe}_{2} \textrm{O}_{3}(s)\) ΔH = -1652 kJ
02

Use stoichiometry to find the heat released

According to the balanced equation, 4 moles of Fe react to produce 2 moles of Fe₂O₃, releasing -1652 kJ of heat. When 4 moles of Fe are reacted, the heat released is the same as the enthalpy change: Heat released = -1652 kJ b. How much heat is released when 1.00 mole of \(\mathrm{Fe}_{2}\mathrm{O}_{3}\) is produced?
03

Use stoichiometry to find the heat released

According to the balanced equation, when 2 moles of Fe₂O₃ are produced, -1652 kJ of heat is released. To find the heat released when 1 mole of Fe₂O₃ is produced, simply divide the enthalpy change by 2: Heat released = -1652 kJ / 2 = -826 kJ c. How much heat is released when \(1.00 \mathrm{g}\) iron is reacted with excess \(\mathbf{O}_{2} ?\)
04

Convert grams of iron to moles

To convert grams to moles, we will use the molar mass of iron(Fe): 55.845 g/mol. Moles of Fe = 1.00 g / 55.845 g/mol = 0.0179 mol
05

Use stoichiometry to find the heat released

The balanced equation tells us that 4 moles of Fe reacts and releases -1652 kJ of heat. We will find the heat released for 0.0179 moles of Fe. Heat released = (0.0179 mol Fe) * (-1652 kJ / 4 mol Fe) = -7.49 kJ (rounded to two decimal places) d. How much heat is released when \(10.0 \mathrm{g}\) Fe and \(2.00 \mathrm{g}\mathrm{O}_{2}\) are reacted?
06

Convert grams of Fe and O₂ to moles

Using the molar mass of Fe(55.845 g/mol) and O₂(32.00 g/mol), convert the grams to moles. Moles of Fe = 10.0 g / 55.845 g/mol = 0.179 mol Moles of O₂ = 2.00 g / 32.00 g/mol = 0.0625 mol
07

Determine the limiting reactant

Based on the balanced equation, we know that 4 moles of Fe react with 3 moles of O₂. We can find the mole ratios of Fe and O₂ in the reaction: Mole ratio (Fe:O₂) = (0.179 mol / 4) / (0.0625 mol / 3) = 1.34 Since this ratio is greater than 1, O₂ is the limiting reactant.
08

Use stoichiometry to find the heat released

Based on the balanced equation and knowing that O₂ is the limiting reactant, we can find the heat released for the given amount of reactants: Heat released = (0.0625 mol O₂) * (-1652 kJ / (3 mol O₂)) = -34.39 kJ (rounded to two decimal places)

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Most popular questions from this chapter

The enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(g)\) when \(\mathrm{H}_{2} \mathrm{O}(g)\) is formed is \(-803 \mathrm{kJ} / \mathrm{mol} .\) Use these data and Hess's law to determine the enthalpy of vaporization for water.

Calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ given the following data: $$\begin{array}{lr}\text { Equation } & \Delta H(\mathrm{kJ}) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) & -1010 \\ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) & -143 \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & -286 \end{array}$$

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