Problem 78 - General Chemistry: Principles and Modern Applications

Question

How many grams of \(\mathrm{CaH}_{2}(\mathrm{s})\) are required to generate sufficient \(\mathrm{H}_{2}(\mathrm{g})\) to fill a \(235 \mathrm{L}\) weather observation balloon at \(722 \mathrm{mmHg}\) and \(19.7^{\circ} \mathrm{C} ?\) \(\mathrm{CaH}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})+2 \mathrm{H}_{2}(\mathrm{g})\).

Step-by-Step Solution

Answer

Approximately 191.5 grams of \(CaH2\) is required.

Step 1: Convert given conditions to moles of \(H2\)

First, convert the given conditions to moles of \(H2\) using the Ideal Gas Law, \(PV = nRT\). Here, pressure P = 722mmHg which is 0.95 atmospheres, V is volume = 235L, R is the gas constant =0.0821 L.atm/K.mol, and T is the temperature = 19.7°C which is 292.85 K when converted. Solve for n, the number of moles of \(H_2\), thus: \(n = PV/RT = (0.95 * 235) / (0.0821 * 292.85) ≈ 9.1 mol\).

Step 2: Convert moles of \(H2\) to moles of \(CaH2\)

Once the number of moles of \(H2\) is found, convert the moles of \(H2\) to moles of \(CaH2\) by the stoichiometry of the balanced equation. This is done because for every 1 mole of \(CaH2\), 2 moles of \(H2\) gas are produced. Therefore, the number of moles of \(CaH2\) = 9.1 mol / 2 ≈ 4.55 mol.

Step 3: Convert moles to grams

The final step is to convert moles of \(CaH2\) to grams. Using the molecular weight of \(CaH2\) which is 42.094 g/mol, the mass of \(CaH2\) required = 4.55 mol * 42.094 g/mol ≈ 191.5 g.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a powerful equation used to relate the properties of an ideal gas. It is expressed as \(PV = nRT\), where:

• \(P\) stands for pressure
• \(V\) indicates volume
• \(n\) represents the number of moles
• \(R\) is the gas constant
• \(T\) signifies temperature

This law is fundamental in solving problems involving gases, as it helps us relate pressure, volume, and temperature to the amount of gas in moles.
In the given exercise, we use the Ideal Gas Law to find the number of moles of hydrogen gas, \(H_2\), required to fill a balloon. By rearranging the equation to solve for \(n\) (\(n = \frac{PV}{RT}\)), we determine how much hydrogen is needed under the given conditions of pressure and temperature.

Moles

Moles are a unit of measurement in chemistry, used to express amounts of a chemical substance. One mole contains exactly \(6.022 \times 10^{23}\) particles, which could be atoms, molecules, or ions. This large number is known as Avogadro's number.
In stoichiometry, moles make it easier to calculate the proportional amounts of reactants and products in chemical reactions.
In our scenario, we calculated that approximately \(9.1\) moles of \(H_2\) gas are required for the balloon. By using a balanced chemical equation, we can convert these moles into an equivalent number of moles of \(\mathrm{CaH}_{2}\).

Chemical Reactions

A chemical reaction involves the transformation of reactants into products through breaking and forming chemical bonds. A balanced chemical equation shows equal numbers of each type of atom on both sides of the equation. In the problem, the reaction given is:
\[\mathrm{CaH}_{2} (s) + 2 \mathrm{H}_{2} \mathrm{O} (l) \rightarrow \mathrm{Ca} (\mathrm{OH})_{2} (aq) + 2 \mathrm{H}_{2} (g)\]
This balanced equation highlights that one mole of \(\mathrm{CaH}_{2}\) produces two moles of \(\mathrm{H}_{2}\).
Thus, by knowing how many moles of \(H_2\) are needed, we can determine the moles of \(\mathrm{CaH}_{2}\) needed to produce that amount, making stoichiometry a critical tool for chemists.

Molecular Weight

Molecular weight, or molar mass, is the mass of one mole of a substance, measured in grams per mole (g/mol). It is calculated by summing the atomic weights of all the atoms in a molecule.

• For \(\mathrm{CaH}_{2}\), the molecular weight is obtained by adding the atomic masses of calcium (Ca) and hydrogen (H):
• Calcium (Ca): 40.08 g/mol
• Hydrogen (H): 1.008 g/mol \( \times 2 = 2.016 g/mol\)
• Total: 40.08 + 2.016 = 42.094 g/mol

This value is crucial for converting between moles and grams.
In this problem, after finding the moles of \(\mathrm{CaH}_{2}\) needed, we multiply by its molecular weight to find the mass required to generate the specified amount of hydrogen gas.