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How much heat, in kilojoules, is associated with the production of \(283 \mathrm{kg}\) of slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) $$\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) \quad \Delta H^{\circ}=-65.2 \mathrm{kJ}$$

Short Answer

Expert verified
The total heat energy released in joules. Remember that if the answer is negative, this means that energy is released.

Step by step solution

01

Calculating the number of moles

Firstly, we need to calculate the number of moles in 283 kg of \( \mathrm{Ca}(\mathrm{OH})_2 \) by using the molar mass of \( \mathrm{Ca}(\mathrm{OH})_2 \), which is 74.093 g/mol. Therefore, the number of moles (n) can be calculated by using the formula \( n = \frac{mass}{molar mass} \), where mass = 283000 g (since 1 kg = 1000 g).
02

Determining the total heat of reaction

Now that we have the number of moles of \( \mathrm{Ca}(\mathrm{OH})_2 \), we can calculate the total heat energy released. Given that \( \Delta H = -65.2 \) kJ/mol, the total heat energy (Q) can be calculated by using the formula \( Q = n \Delta H \).
03

Calculating the total heat energy

Substitute the values into the formula. The sign would remain negative as the heat is released and not absorbed. This will give you the final answer in kilojoules.

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is an important concept in understanding how heat is transferred in chemical reactions. It signifies the difference in enthalpy—essentially, the total heat content—between the products and the reactants. Whether heat is absorbed or released during the reaction is determined by the sign of \( \Delta H \).

  • Exothermic reactions—like the one producing slaked lime, \( \text{Ca(OH)}_2 \)—release heat, signified by a negative \( \Delta H \).
  • Endothermic reactions absorb heat, indicated by a positive \( \Delta H \).

To calculate the total heat produced in a reaction, like when producing slaked lime, we multiply the number of moles of the product by the standard enthalpy change per mole. This gives the total amount of heat energy involved, expressed in kilojoules.
Molar Mass
Molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). Understanding this property is essential for converting between the mass of a substance and its amount in moles, facilitating stoichiometric calculations.

To find the molar mass, each element's atomic mass (from the periodic table) within a compound is used. For \( \text{Ca(OH)}_2 \):
  • Calcium (Ca): Approx. 40.08 g/mol
  • Oxygen (O): Approx. 16.00 g/mol
  • Hydrogen (H): Approx. 1.01 g/mol
    • Add the contributions together:
      • Calcium: \( 1 \times 40.08 \text{ g/mol} \)
      • Oxygen: \( 2 \times 16.00 \text{ g/mol} \)
      • Hydrogen: \( 2 \times 1.01 \text{ g/mol} \)
      Totaling: \( 40.08 + 32.00 + 2.02 = 74.10 \text{ g/mol} \)
      Therefore, for \( 283 \text{ kg} \) of slaked lime equivalent to \( 283000 \text{ g} \), the number of moles is calculated using:

      \[ n = \frac{283000 \text{ g}}{74.10 \text{ g/mol}} \]
Stoichiometry
Stoichiometry involves using balanced chemical equations to relate the amounts of reactants and products in a reaction. It is hugely valuable in predicting the outcomes of chemical reactions based on known quantities.

In the chemical reaction to produce slaked lime, we have:
  • Reactants: Calcium oxide \( \text{CaO} \) and Water \( \text{H}_2\text{O} \)
  • Product: Slaked lime \( \text{Ca(OH)}_2 \)
The balanced equation, \[ \text{CaO(s)} + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(s) \], indicates that one mole of calcium oxide reacts with one mole of water to produce one mole of slaked lime.

For the given problem, after determining the moles of \( \text{Ca(OH)}_2 \) produced, the stoichiometric relation tells us this same number of moles corresponds to the initial quantities of reactants. Using the number of moles and the known \( \Delta H \), we calculate the total heat released, which is critical in many practical applications, like processing reactions on an industrial scale.

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Most popular questions from this chapter

We can determine the purity of solid materials by using calorimetry. A gold ring (for pure gold, specific heat \(=0.1291 \mathrm{Jg}^{-1} \mathrm{K}^{-1}\) ) with mass of \(10.5 \mathrm{g}\) is heated to \(78.3^{\circ} \mathrm{C}\) and immersed in \(50.0 \mathrm{g}\) of \(23.7^{\circ} \mathrm{C}\) water in a constant-pressure calorimeter. The final temperature of the water is \(31.0^{\circ} \mathrm{C}\). Is this a pure sample of gold?

What will be the final temperature of the water in an insulated container as the result of passing \(5.00 \mathrm{g}\) of steam, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) at \(100.0^{\circ} \mathrm{C}\) into \(100.0 \mathrm{g}\) of water at \(25.0^{\circ} \mathrm{C} ?\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

A 1.103 g sample of a gaseous carbon-hydrogenoxygen compound that occupies a volume of \(582 \mathrm{mL}\) at 765.5 Torr and \(25.00^{\circ} \mathrm{C}\) is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb calorimeter. The products of the combustion are \(2.108 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}), 1.294 \mathrm{g} \mathrm{H}_{2} \mathrm{O}(1),\) and enough heat to raise the temperature of the calorimeter assembly from 25.00 to \(31.94^{\circ} \mathrm{C}\). The heat capacity of the calorimeter is \(5.015 \mathrm{kJ} /^{\circ} \mathrm{C}\). Write an equation for the combustion reaction, and indicate \(\Delta H^{\circ}\) for this reaction at \(25.00^{\circ} \mathrm{C}\).

A 75.0 g piece of \(\mathrm{Ag}\) metal is heated to \(80.0^{\circ} \mathrm{C}\) and dropped into \(50.0 \mathrm{g}\) of water at \(23.2^{\circ} \mathrm{C} .\) The final temperature of the \(\mathrm{Ag}-\mathrm{H}_{2} \mathrm{O}\) mixture is \(27.6^{\circ} \mathrm{C}\). What is the specific heat of silver?

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

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