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Chapter 10: Problem 122

A 4.00-g sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a \(1.00-\mathrm{L}\) vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO}\), forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3}\). When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

Short Answer

Expert verified
0.0313 moles of CO鈧� have reacted, and the mass percentage of CaO in the mixture is 10.30%.

Step by step solution

01

Find the initial moles of CO鈧�

To find the initial moles of CO鈧�, we can use the ideal gas law: \(PV = nRT\), where P is pressure, V is volume, n is moles, R is the gas constant, and T is temperature in Kelvin. Rearrange the formula to solve for n: \(n = \frac{PV}{RT}\) The initial pressure of CO鈧� is given as 730 torr. To convert this to atmospheres, divide by 760 torr/atm: \(730 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 0.9605 \text{ atm}\) Now we can plug in the values: - P = 0.9605 atm - V = 1.00 L - R = 0.0821 \(\frac{\text{L atm}}{\text{mol K}}\) (Ideal gas constant) - T = 25掳C + 273.15 = 298.15 K \[n_\text{initial} = \frac{(0.9605 \text{ atm})(1.00 \text{ L})}{(0.0821 \frac{\text{L atm}}{\text{mol K}})(298.15 \text{ K})} = 0.0394 \text{ moles of CO}_{2}\]
02

Find the final moles of CO鈧�

After the reaction, the pressure of CO鈧� is given as 150 torr. Convert to atmospheres: \(150 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 0.1974 \text{ atm}\) Using the same method as before, plug in the values to the ideal gas law equation: - P = 0.1974 atm - V = 1.00 L - R = 0.0821 \(\frac{\text{L atm}}{\text{mol K}}\) - T = 298.15 K \[n_\text{final} = \frac{(0.1974 \text{ atm})(1.00 \text{ L})}{(0.0821 \frac{\text{L atm}}{\text{mol K}})(298.15 \text{ K})} = 0.0081 \text{ moles of CO}_{2}\]
03

Calculate the number of moles of CO鈧� that have reacted

To find the moles of CO鈧� that have reacted, subtract the final moles of CO鈧� from the initial moles of CO鈧�: \[n_\text{reacted} = n_\text{initial} - n_\text{final} = 0.0394 - 0.0081 = 0.0313 \text{ moles of CO}_{2}\] So, 0.0313 moles of CO鈧� have reacted.
04

Calculate the moles of CaO and BaO in the mixture

In the reaction, CO鈧� reacts with CaO and BaO to form CaCO鈧� and BaCO鈧�. The balanced equations are: \[1)\,\text{CaO} + \text{CO}_{2} 鈫� \text{CaCO}_{3}\] \[2)\,\text{BaO} + \text{CO}_{2} 鈫� \text{BaCO}_{3}\] Since both reactions consume 1 mole of CO鈧� per 1 mole of CaO/BaO, the total moles of CaO and BaO in the mixture are equal to the moles of CO鈧� reacted: 0.0313 moles.
05

Calculate the mass percentage of CaO in the mixture

Let x = moles of CaO in the mixture. Then, the moles of BaO in the mixture will be 0.0313 - x. Next, convert moles to mass using molar mass: Mass of CaO (g) = x moles 脳 56.08 g/mol (Molar mass of CaO = 56.08 g/mol) Mass of BaO (g) = (0.0313 - x) moles 脳 153.33 g/mol (Molar mass of BaO = 153.33 g/mol) The sum of the masses of CaO and BaO should be equal to the total mass of the mixture (4.00 g). This gives us the equation: \(56.08x + 153.33(0.0313 - x) = 4.00\) Now, solve the equation for x: \[\begin{aligned} 56.08x + 153.33(0.0313 - x) &= 4.00 \\ 56.08x + 4.8003 - 153.33x &= 4.00 \\ -97.25x &= -0.8003 \\ x &= 0.00823 \end{aligned}\] Now we can calculate the mass percentage of CaO: \[\frac{\text{mass of CaO}}{\text{total mass}} = \frac{(0.00823 \text{ moles})(56.08\frac{\text{g}}{\text{mol}})}{4.00\text{ g}}\times 100\%= 10.30\%\] So, the mass percentage of CaO in the mixture is 10.30%.

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Key Concepts

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

Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. Given by the formula \(PV = nRT\), where \(P\) denotes pressure, \(V\) is volume, \(n\) stands for moles, \(R\) represents the ideal gas constant (0.0821 L atm/mol K), and \(T\) is the temperature in Kelvin.

For solving problems involving the ideal gas law, it's vital to ensure that the pressure is in atmospheres and the temperature is in Kelvin. When given other units, such as torr for pressure, a conversion is necessary (1 atm = 760 torr). In the provided exercise, the initial and final pressures of the \(CO_2\) gas were converted from torr to atmospheres to solve for the moles before and after the reaction.

Moreover, the ideal gas law assumes that gases behave ideally, which means gas particles do not attract or repel each other and don't occupy space. Although real gases do not perfectly comply with these assumptions, the ideal gas law offers a close approximation for many conditions, particularly under standard temperature and pressure.
Mole Concept
The mole concept is central to chemistry as it provides a bridge between the atomic scale and the macroscopic world. One mole is defined as exactly \(6.022 \times 10^{23}\) particles (Avogadro's number) of the substance, which could be atoms, molecules, ions, or other entities depending on the context.

In relation to the exercise, understanding the mole concept is essential for converting between moles and grams. By knowing the molar mass of a substance (in grams per mole), we can calculate how many moles are present in a given mass or conversely, how many grams are represented by a certain number of moles. This is pivotal when computing the amount of \(CO_2\) that has reacted, and subsequently, the individual amounts of \(CaO\) and \(BaO\) present in the initial sample.

Why is understanding the mole concept important?

Knowing the mole concept allows chemists to count particles by weighing, which is practical and facilitates stoichiometric calculations in chemical reactions.
Percent Composition
Percent composition refers to the mass percent of each element in a compound or a mixture. It's calculated by dividing the mass of a particular component by the total mass and then multiplying by 100%. This concept is vital when dealing with mixtures or compounds as it leads to an understanding of the composition and can aid in determining empirical and molecular formulas.

In the problem at hand, after determining the number of moles of \(CaO\) and \(BaO\) that reacted with \(CO_2\), the mass percentage of \(CaO\) in the original mixture could be calculated. You find the mass of \(CaO\) by multiplying the moles of \(CaO\) by its molar mass. The total mass of the mixture is known (4.00 g), allowing the determination of the mass percentage of \(CaO\) after solving the algebraic equations obtained from the stoichiometry of the reactions.

Practical Applications of Percent Composition

Knowledge of percent composition is essential not only for chemical synthesis but also for quality control in various industries, such as pharmaceuticals and food manufacturing, where the components' proportions can affect product efficacy and safety.

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Most popular questions from this chapter

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\) ? (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL}\); the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

The temperature of a \(5.00-\mathrm{L}\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root- meansquare speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls per second.

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(C\), is added to the same vessel at the same temperature. How does the addition of gas \(C\) affect the following: (a) the partial pressure of gas A, (b) the total pressure in the vessel, (c) the mole fraction of gas B?

Assume that a single cylinder of an automobile engine has a volume of \(524 \mathrm{~cm}^{3}\). (a) If the cylinder is full of air at \(74{ }^{\circ} \mathrm{C}\) and \(0.980 \mathrm{~atm}\), how many moles of \(\mathrm{O}_{2}\) are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095\).) (b) How many grams of \(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? as originally filled with gas?

In the United States, barometric pressures are generally reported in inches of mercury (in. \(\mathrm{Hg}\) ). On a beautiful summer day in Chicago, the barometric pressure is \(30.45 \mathrm{in} . \mathrm{Hg}\). (a) Convert this pressure to torr. (b) Convert this pressure to atm.
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