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Chapter 10: Problem 59

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)$$ If 159 \(\mathrm{mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of Zn have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Short Answer

Expert verified
0.402 grams of zinc have been consumed in the reaction.

Step by step solution

01

(Step 1) Calculate moles of hydrogen gas collected.

First, we need to find the moles of hydrogen gas collected. Since we are given the volume, temperature, and pressure of the hydrogen, we can use the ideal gas law to find the amount (in moles) of hydrogen gas. The ideal gas law is given by: PV = nRT. Where: - P is pressure (in atm) - V is volume (in L) - n is the amount (in moles) - R is the ideal gas constant (0.08206 L atm/K mol) - T is the temperature (in K) First, convert the given data to the appropriate units: - Volume: 159 mL to 0.159 L (by dividing by 1000) - Temperature: 24°C to 297 K (by adding 273) - Pressure: 738 torr to \(0.970 \,\mathrm{atm}\) (by dividing by 760) Given the vapor pressure of water, we need to find the pressure of only the hydrogen gas. Since the hydrogen gas is collected over water, the total pressure is the sum of the hydrogen gas pressure and the vapor pressure of water. We can find the vapor pressure of water at 24°C in the Appendix B, which is approximately 22.4 torr. Convert this value to atm: \(\frac{22.4 \,\mathrm{torr}}{760 \,\mathrm{torr/atm}} = 0.0295 \,\mathrm{atm}\). Now, find the pressure of the hydrogen gas by subtracting the vapor pressure of water from the barometric pressure: \(0.970 \,\mathrm{atm} - 0.0295 \,\mathrm{atm} = 0.9405 \,\mathrm{atm}\). Now use the ideal gas law to solve for the moles of hydrogen gas (n): n = \(\frac{PV}{RT}\). n(H₂) = \(\frac{(0.9405 \,\mathrm{atm})(0.159 \,\mathrm{L})}{(0.08206 \,\mathrm{L\, atm/K\, mol})(297\,\mathrm{K})} = 0.00616 \,\mathrm{mol}\).
02

(Step 2) Determine moles of consumed zinc.

According to the balanced chemical equation, the mole ratio of Zn to H₂ is 1:1, which means that for every mole of hydrogen gas produced, one mole of zinc is consumed. So the moles of consumed zinc (Zn) is the same as the moles of hydrogen gas collected: n(Zn) = n(H₂) = 0.00616 mol
03

(Step 3) Calculate the mass of consumed zinc.

Now that we have the moles of consumed zinc, we can find the mass (in grams) by multiplying the moles by the molar mass of zinc (65.38 g/mol): Mass(Zn) = n(Zn) × Molar mass(Zn) Mass(Zn) = (0.00616 mol)(65.38 g/mol) = 0.402 g So, 0.402 grams of zinc have been consumed in the reaction.

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Key Concepts

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

Understanding the Ideal Gas Law
When studying chemical reactions involving gases, the ideal gas law is an essential tool for finding relationships between pressure, volume, temperature, and the number of moles of a gas. The ideal gas law formula is expressed as PV = nRT, where P stands for pressure, V represents volume, n indicates the amount of gas in moles, R is the universal gas constant, and T is the absolute temperature in Kelvin.

To apply this law, it's necessary to match the units of the constant R with the units used for pressure and volume. In the context of the textbook problem, the pressure was converted to atmospheres (atm) and the volume was converted to liters (L) to align with the value of R (0.08206 L atm/K mol). Temperature must always be in Kelvin, which requires adding 273 to the Celsius scale.

Furthermore, when dealing with gases collected over water, such as hydrogen in the exercise, it's crucial to account for the partial pressure of water vapor. This detail involves subtracting the vapor pressure of water from the total pressure to find the pressure exerted solely by the hydrogen gas. This corrected pressure is then used in the ideal gas law equation to solve for the number of moles of gas collected.

By mastering the usage of the ideal gas law, students can confidently solve problems dealing with gaseous reactions under various conditions, ensuring accurate stoichiometric calculations.
The Process of Molar Mass Calculation
Molar mass is a critical concept in chemistry, indicating the mass of one mole of a substance. This property bridges the gap between the macroscopic scale of grams and the microscopic scale of atoms and molecules. In stoichiometry, the molar mass calculation allows us to convert between mass and moles, which is fundamental for quantifying substances in a reaction.

The molar mass of an element can be found on the periodic table as the atomic weight, typically given in grams per mole (g/mol). For compounds, the molar mass is the sum of the atomic weights of all atoms in the formula. In the exercise, the molar mass of zinc (Zn) was necessary to calculate the mass of zinc that reacted. After obtaining the moles of zinc from Step 2, the calculation was straightforward: just multiplying the molar mass of zinc by the number of moles.

This simple multiplication provided us with the mass in grams of zinc that had been consumed: Mass(Zn) = (Moles of Zn) × (Molar mass of Zn). This step underscores the importance of molar mass as a conversion factor in stoichiometry, as it directly relates to the amount of a substance in a chemical reaction.
Chemical Reaction Balancing Essentials
Properly balancing a chemical equation is the foundational step in stoichiometry, ensuring that the principle of conservation of mass is upheld. In chemical reactions, atoms are neither created nor destroyed, so we must have the same number and types of atoms on both sides of the reaction. This balance allows us to calculate reactants and products accurately.

The given exercise presents the reaction between zinc and sulfuric acid, producing zinc sulfate and hydrogen gas. To balance the equation, we need to verify that the number of atoms for each element is equal on the reactants and products sides. In this case, the reaction is already balanced, with one zinc atom, one sulfur atom, four oxygen atoms, and two hydrogen atoms on both sides.

Understanding the stoichiometry of the reaction, which in this scenario is a 1:1 ratio between zinc and hydrogen gas, allows us to use the moles of one substance to find the moles of another. This equivalency is crucial for solving multi-step problems that require the conversion from one substance to another, as seen in this textbook example. By emphasizing the necessity to balance equations and molar ratios, we set up a clear pathway to solving quantitative problems in chemistry.

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Most popular questions from this chapter

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

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 are high because it is necessary to point at atmospheric pressure of \(-164^{\circ} \mathrm{C} .\) One possible strategy is to oxidize the methane to methanol, 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?

A 15.0 -L tank is filled with helium gas at a pressure of \(1.00 \times 10^{2}\) atm. How many balloons (each 2.00 L) can be inflated to a pressure of 1.00 atm, assuming that the temperature remains constant and that the tank cannot be emptied below 1.00 atm?

Carbon dioxide, which is recognized as the major contributor to global warming as a "greenhouse gas," is formed when fossil fuels are combusted, as in electrical power plants fueled by coal, oil, or natural gas. One potential way to reduce the amount of \(\mathrm{CO}_{2}\) added to the atmosphere is to store it as a compressed gas in underground formations.Consider a 1000 -megawatt coal-fired power plant that produces about \(6 \times 10^{6}\) tons of \(\mathrm{CO}_{2}\) per year. (a) Assuming ideal-gas behavior, 1.00 atm, and \(27^{\circ} \mathrm{C},\) calculate the volume of \(\mathrm{CO}_{2}\) produced by this power plant. (b) If the \(\mathrm{CO}_{2}\) is stored underground as a liquid at \(10^{\circ} \mathrm{C}\) and 120 \(\mathrm{atm}\) and a density of \(1.2 \mathrm{g} / \mathrm{cm}^{3},\) what volume does it possess?(c) If it is stored underground as a gas at \(30^{\circ} \mathrm{C}\) and \(70 \mathrm{atm},\) what volume does it occupy?

A 35.1 g sample of solid \(\mathrm{CO}_{2}(\) dry ice \()\) is added to a container at a temperature of 100 \(\mathrm{K}\) with a volume of 4.0 \(\mathrm{L} .\) If the container is evacuated (all of the gas removed), sealed and then allowed to warm to room temperature \((T=298 \mathrm{K})\) so that all of the solid \(\mathrm{CO}_{2}\) is converted to a gas, what is the pressure inside the container?
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