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Chapter 6: Problem 49

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

Expert verified
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 O₂ are reacted?
07

Calculate the moles of Fe and O₂

Calculate the number of moles of Fe and O₂ involved: \[ \text{Moles of Fe} = \frac{10.0 \, \text{g}}{55.85 \, \text{g/mol}} = 0.179 \, \text{mol} \] \[ \text{Moles of O₂} = \frac{2.00 \, \text{g}}{32.00 \, \text{g/mol}} = 0.0625 \, \text{mol} \]
08

Identify the limiting reactant

Determine the limiting reactant by comparing the moles of Fe and O₂ with their stoichiometric coefficients (4 for Fe and 3 for O₂): \[ \frac{\text{moles of Fe}}{4} = \frac{0.179 \, \text{mol}}{4} = 0.0448 \, \text{mol} \] \[ \frac{\text{moles of O₂}}{3} = \frac{0.0625 \, \text{mol}}{3} = 0.0208 \, \text{mol} \] Since 0.0208 < 0.0448, O₂ is the limiting reactant.
09

Calculate the heat released

Now we can determine the heat released when 0.0625 moles of O₂ react with excess Fe, using the stoichiometry ratio and the heat released by 3 moles of O₂: \[ \text{Heat released} = \frac{0.0625 \, \text{mol}}{3.00 \, \text{mol}} \times -1652 \, \text{kJ} = -34.3 \, \text{kJ} \]

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Most popular questions from this chapter

At 298 \(\mathrm{K}\) , the standard enthalpies of formation for $\mathrm{C}_{2} \mathrm{H}_{2}(g)$ and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are 227 \(\mathrm{kJ} / \mathrm{mol}\) and \(49 \mathrm{kJ} / \mathrm{mol},\) respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

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?

A 5.00 -g sample of aluminum pellets (specific heat capacity \(=\) 0.89 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) ) and a \(10.00-\mathrm{g}\) sample of iron pellets (specific heat capacity $=0.45 \mathrm{J} / \mathrm{C} \cdot \mathrm{g}\( are heated to \)100.0^{\circ} \mathrm{C}$ . The mixture of hot iron and aluminum is then dropped into 97.3 \(\mathrm{g}\) water at $22.0^{\circ} \mathrm{C}$ . Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
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