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Chapter 10: Problem 115

A \(4.00-\mathrm{g}\) sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-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
The number of moles of CO鈧� that have reacted is 0.03135 mol. The mass percentage of CaO in the mixture is 33.28%.

Step by step solution

01

Find the initial number of moles of CO鈧�

Start by finding the number of moles of CO鈧� at the beginning, using the Ideal Gas Law, which states: PV = nRT Where - P is the pressure (converted to atm) - V is the volume - n is the number of moles - R is the gas constant (0.08206 L atm/mol K) - T is the temperature (converted to Kelvin) Initially, the pressure of CO鈧� is 730 torr. First, convert this to atm: \( (730\,\text{torr})(\frac{1\,\text{atm}}{760\,\text{torr}}) = 0.9605\,\text{atm} \). The temperature is given as 25掳C. Convert this to Kelvin: \( 25 + 273.15 = 298.15\,\text{K} \). The volume of the vessel is 1.00 L. Now, apply the Ideal Gas Law to calculate the number of moles of CO鈧�: \( n = \frac{PV}{RT} = \frac{(0.9605\,\text{atm})(1.00\,\text{L})}{(0.08206\,\text{L atm/mol K})(298.15\,\text{K})} = 0.03945\,\text{mol} \)
02

Find the final number of moles of CO鈧� after the reaction

After the reaction, the pressure of CO鈧� is 150 torr, which needs to be converted to atm: \( (150\,\text{torr})(\frac{1\,\text{atm}}{760\,\text{torr}}) = 0.19737\,\text{atm} \). Apply the Ideal Gas Law to calculate the number of moles of CO鈧� after the reaction: \( n = \frac{PV}{RT} = \frac{(0.19737\,\text{atm})(1.00\,\text{L})}{(0.08206\,\text{L atm/mol K})(298.15\,\text{K})} = 0.008097\,\text{mol} \)
03

Calculate the number of moles of CO鈧� reacted

Subtract the final number of moles of CO鈧� from the initial number of moles: \( 0.03945\,\text{mol} - 0.008097\,\text{mol} = 0.03135\,\text{mol CO2 reacted} \)
04

Determine moles of CO鈧� reacted with CaO and BaO separately

Let x be the moles of CO鈧� reacted with CaO; then, (0.03135 - x) is the moles of CO鈧� reacted with BaO. The reaction with CaO is: CaO + CO鈧� 鈫� CaCO鈧�, And the reaction with BaO is: BaO + CO鈧� 鈫� BaCO鈧� Now, calculate the mass of CaO and BaO in the sample using their molar masses: - Mass of CaO = mass of the sample - mass of BaO - Mass of CaO = 4.00 g - mass of BaO - Mass of CaO = x * (molar mass of CaO) = x * 56.079 g/mol - Mass of BaO = (0.03135 - x) * (molar mass of BaO) = (0.03135 - x) * 153.33 g/mol Since Mass of CaO + Mass of BaO = 4.00 g, we can create the following equation: \( x(56.079) + (0.03135 - x)(153.33) = 4.00 \)
05

Calculate the mass percentage of CaO in the mixture

Solve the equation from step 4 for x: \( x = 0.02374\,\text{mol of CO2 reacted with CaO} \) Now, calculate the mass of CaO in the sample: \( \text{mass of CaO} = 0.02374\,\text{mol} * 56.079\,\text{g/mol} = 1.331\,\text{g} \) Finally, calculate the mass percentage of CaO in the mixture: \( \text{mass percentage of CaO} = \frac{1.331\,\text{g}}{4.00\,\text{g}} * 100 = 33.28\% \) Thus, the mass percentage of CaO in the mixture is 33.28%.

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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, advanced by combining several empirical gas laws. It is expressed as \( PV = nRT \), where:
  • P is the pressure of the gas in atmospheres (atm).
  • V is the volume of the gas in liters (L).
  • n is the number of moles of the gas.
  • R is the universal gas constant (0.0821 L atm/mol K).
  • T is the temperature in Kelvin (K).
The Ideal Gas Law ties together pressure, volume, temperature, and the amount of gas into one coherent equation. This equation is incredibly useful for calculating missing variables when you know the other three. In our exercise, we used this formula to determine the moles of \( \mathrm{CO}_{2} \) both initially and after the reaction occurred. Start by converting given units to the ideal units: convert pressure from torr to atm by dividing by 760, and temperature from Celsius to Kelvin by adding 273.15. Once everything is set, plug the values into the equation and solve for the unknown!
Reaction Calculations
Reaction calculations are crucial in stoichiometry as they help determine how reactants convert to products. It involves several steps:
Step 1: Calculate Initial and Remaining Moles
Before and after the reaction, the moles of \( \mathrm{CO}_{2} \) were calculated using the Ideal Gas Law. This gave us an understanding of the amount of \( \mathrm{CO}_{2} \) present initially and after forming calcium carbonate and barium carbonate.
Step 2: Determine Moles Reacted
By taking the difference between initial and remaining moles of \( \mathrm{CO}_{2} \), we derived the moles of \( \mathrm{CO}_{2} \) that reacted with \( \mathrm{CaO} \) and \( \mathrm{BaO} \).
Step 3: Identify Individual Reactions
Knowing the chemical equations \( \mathrm{CaO} + \mathrm{CO}_{2} \rightarrow \mathrm{CaCO}_{3} \) and \( \mathrm{BaO} + \mathrm{CO}_{2} \rightarrow \mathrm{BaCO}_{3} \), helps us understand complete conversion, which is crucial for further stoichiometric calculations like determining masses.
Gas Pressure Conversion
Gas pressure conversion is often necessary in chemical calculations to ensure all units are compatible, particularly when using the Ideal Gas Law. Pressure in atmospheres (atm) is standard for the law, but many situations, such as this exercise, present pressure in torr. Conversion is straightforward: 1 atm is equivalent to 760 torr.
Example Conversion
Given the initial pressure of \( \mathrm{CO}_{2} \) was 730 torr, it needed to be converted to atm to fit the Ideal Gas Law. Use the relation:\[\text{Pressure in atm} = \frac{\text{Pressure in torr}}{760}\]For 730 torr, this comes out to \(0.9605\, \text{atm}\). Similarly, for the final pressure of 150 torr, the conversion yields \(0.19737\, \text{atm}\). These conversions were pivotal in correctly applying the Ideal Gas Law to determine moles before and after the reaction, thus ensuring accuracy in our stoichiometric calculations.

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Most popular questions from this chapter

A sample of \(3.00 \mathrm{~g}\) of \(\mathrm{SO}_{2}(\mathrm{~g})\) originally in a \(5.00\) -L vessel at \(21^{\circ} \mathrm{C}\) is transferred to a \(10.0\) -L vessel at \(26^{\circ} \mathrm{C}\). A sample of \(2.35 \mathrm{~g} \mathrm{~N}_{2}(g)\) originally in a \(2.50\) -L vessel at \(20^{\circ} \mathrm{C}\) is transferred to this same \(10.0\) - \(\mathrm{L}\) vessel. (a) What is the partial pressure of \(\mathrm{SO}_{2}(g)\) in the larger container? (b) What is the partial pressure of \(\mathrm{N}_{2}(g)\) in this vessel? (c) What is the total pressure in the vessel?

The molar mass of a volatile substance was determined by the Dumas-bulb method described in Exercise \(10.51\). The unknown vapor had a mass of \(0.846 \mathrm{~g} ;\) the volume of the bulb was \(354 \mathrm{~cm}^{3}\), pressure 752 torr, and temperature \(100^{\circ} \mathrm{C}\). Calculate the molar mass of the unknown vapor.

A gas bubble with a volume of \(1.0 \mathrm{~mm}^{3}\) originates at the bottom of a lake where the pressure is \(3.0 \mathrm{~atm}\). Calculate its volume when the bubble reaches the surface of the lake where the pressure is 695 torr, assuming that the temperature doesn't change.

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 thus 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 \(25^{\circ} \mathrm{C}\) are 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 described above 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?

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