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Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate \(145 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is 825 torr at \(21^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
To generate 145 L of hydrogen gas at a pressure of 825 torr and a temperature of 21°C, it is required to use 142.04 grams of calcium hydride (\(\mathrm{CaH}_{2}\)).

Step by step solution

01

Convert the given Data to Required Units

First, let's convert the given pressure and temperature to standard units. Pressure in atm and temperature in Kelvin. Pressure, P (in atm) = 825 torr × (1 atm / 760 torr) = 1.0855 atm Temperature, T (in K) = 21°C + 273.15 = 294.15 K
02

Apply the Ideal Gas Law Equation

The ideal gas law equation is as follows: PV = nRT Where: P = pressure (in atm) V = volume (in L) n = number of moles of gas R = ideal gas constant (0.0821 L atm / mol K) T = temperature (in K) Given volume, V = 145 L, Pressure, P = 1.0855 atm, and Temperature, T = 294.15 K, we can now solve for the number of moles of hydrogen gas (H2) as shown below: n = PV / RT
03

Calculate Number of Moles of Hydrogen Gas

Plug the given data into the equation: n = (1.0855 atm × 145 L) / (0.0821 L atm / mol K × 294.15 K) n = 6.7463 mol We have 6.7463 moles of hydrogen gas (H2).
04

Stoichiometric Calculations

From the balanced chemical equation, \(\mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g)\), we can infer the stoichiometric ratio between \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2}\) as 1:2. Now, we can calculate the moles of \(\mathrm{CaH}_{2}\) required to produce 6.7463 moles of \(\mathrm{H}_{2}\) using the stoichiometric ratio: moles of \(\mathrm{CaH}_{2}\) = (moles of \(\mathrm{H}_{2}\)) × (1 mol \(\mathrm{CaH}_{2}\) / 2 mol \(\mathrm{H}_{2}\)) moles of \(\mathrm{CaH}_{2}\) = 6.7463 mol × (1/2) moles of \(\mathrm{CaH}_{2}\) = 3.37315 mol
05

Convert Moles of \(\mathrm{CaH}_{2}\) to Grams

Finally, convert the moles of calcium hydride (CaH2) to grams using its molar mass. The molar mass of \(\mathrm{CaH}_{2}\) is 42.094 g/mol. grams of \(\mathrm{CaH}_{2}\) = (moles of \(\mathrm{CaH}_{2}\)) × (molar mass of \(\mathrm{CaH}_{2}\)) grams of \(\mathrm{CaH}_{2}\) = 3.37315 mol × 42.094 g/mol grams of \(\mathrm{CaH}_{2}\) = 142.04 g Hence, 142.04 grams of calcium hydride (\(\mathrm{CaH}_{2}\)) are needed to generate 145 L of hydrogen gas at a pressure of 825 torr and a temperature of 21°C.

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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 that describes the behavior of gases, allowing us to interrelate the pressure, volume, temperature, and quantity (in moles) of a gas. It is expressed as:
\[ PV = nRT \]
where:
  • \( P \): Pressure in atmospheres (atm)
  • \( V \): Volume in liters (L)
  • \( n \): Number of moles of gas
  • \( R \): Ideal gas constant, approximately 0.0821 L atm/mol K
  • \( T \): Temperature in Kelvin (K)
This equation helps us calculate one of these properties if the others are known.
In the context of our problem, we use the Ideal Gas Law to determine the number of moles of hydrogen gas produced in a reaction. By rearranging the equation to solve for \( n \) (moles), we substitute the known values of pressure, volume, and temperature into the equation.
This straightforward approach provides a clear pathway to relate physical conditions of a gas with its chemical representation.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products. The essence of a balanced chemical reaction is to convey the conservation of mass during the reaction process, with atoms of each element accounted for on both sides of the equation.
In our example, calcium hydride \((\mathrm{CaH}_2)\) reacts with water to produce calcium hydroxide and hydrogen gas. This reaction is represented by the equation:
\[\mathrm{CaH}_{2}(s) + 2 \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(aq)+2 \mathrm{H}_{2}(g)\]
From this balanced equation, we derive the stoichiometric ratios: one mole of \(\mathrm{CaH}_2\) reacts with two moles of \(\mathrm{H}_2\mathrm{O}\), producing two moles of \(\mathrm{H}_2\) gas.
Understanding these ratios is crucial as they serve as the bridge between chemistry and calculations.
It enables us to determine how much reactant is needed to produce a certain amount of product, reinforcing the practical application of stoichiometry.
Molar Mass
Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). It is determined by summing the atomic masses of all atoms in a molecular formula.
For example, the molar mass of calcium hydride (\(\mathrm{CaH}_2\)) is calculated by adding the atomic masses of calcium (approximately 40.08 g/mol) and two hydrogen atoms (approximately 1.01 g/mol each):
  • Molar mass of \(\mathrm{CaH}_2\) = 40.08 + (2 \times 1.01) = 42.10 g/mol
Molar mass serves as a conversion factor between the mass of a compound and moles, enabling chemists to measure quantities needed for reactions.
In our problem, understanding molar mass allows us to convert moles of \(\mathrm{CaH}_2\) calculated from the stoichiometric relationships into grams, which is more practical for actual laboratory use.
Using this conversion effectively bridges theoretical calculations and real-world applications.
Gas Laws
The Ideal Gas Law is just one component of the broader group known as gas laws. These laws describe the behavior and interactions of gases under varying conditions.
Some of the key gas laws include:
  • Boyle's Law: Relates pressure and volume \((P \times V = \, \text{constant})\) indicating that pressure increases as volume decreases when temperature is held constant.
  • Charles's Law: Shows the relationship between volume and temperature \((V/T = \, \text{constant})\) indicating that volume increases with temperature when pressure is constant.
  • Avogadro's Law: Expresses the relationship between volume and moles \((V/n = \, \text{constant})\), suggesting that volume increases with the number of gas molecules at constant pressure and temperature.
Collectively, these laws, along with the Ideal Gas Law, describe how gases will react to changes in environmental conditions, providing a foundational understanding of gas behavior.
For students, these laws illustrate the dynamic nature of gases and are essential in solving problems involving chemical reactions in gaseous states.

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Most popular questions from this chapter

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), can be prepared by the reaction of calcium carbide with water: $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$ Calculate the volume of \(\mathrm{C}_{2} \mathrm{H}_{2}\) that is collected over water at \(23{ }^{\circ} \mathrm{C}\) by reaction of \(1.524 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) if the total pressure of the gas is 753 torr. (The vapor pressure of water is tabulated in Appendix B.)

The metabolic oxidation of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), in our bodies produces \(\mathrm{CO}_{2}\), which is expelled from our lungs as a gas: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) Calculate the volume of dry \(\mathrm{CO}_{2}\) produced at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and \(0.970 \mathrm{~atm}\) when \(24.5 \mathrm{~g}\) of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at \(1.00\) atm and \(298 \mathrm{~K}\), to completely oxidize \(50.0 \mathrm{~g}\) of glucose.

Perform the following conversions: (a) \(0.912\) atm to torr, (b) \(0.685\) bar to kilopascals, (c) \(655 \mathrm{~mm} \mathrm{Hg}\) to atmospheres, (d) \(1.323 \times 10^{5} \mathrm{~Pa}\) to atmospheres, (e) \(2.50 \mathrm{~atm}\) to psi.

(a) What are the mole fractions of each component in a mixture of \(15.08 \mathrm{~g}\) of \(\mathrm{O}_{2}, 8.17 \mathrm{~g}\) of \(\mathrm{N}_{2}\), and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\) ? (b) What is the partial pressure in atm of each component of this mixture if it is held in a 15.50- \(\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C}\) ?

To derive the ideal-gas equation, we assume that the volume of the gas atoms/molecules can be neglected. Given the atomic radius of neon, \(0.69 \AA\), and knowing that a sphere has a volume of \(4 \pi \mathrm{r}^{3} / 3\), calculate the fraction of space that Ne atoms occupy in a sample of neon at STP.

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