Problem 58 - Chemistry The Central Science

Question

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 \(21{ }^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

To generate 145 L of \(\mathrm{H}_{2}\) gas at 825 Torr and 21°C, 143.493 g of \(\mathrm{CaH}_{2}\) are needed.

Step 1: Convert given information

First, we need to convert the given pressure and temperature to common units: Pressure (P): 825 torr = \( \frac{825}{760} \) atm = 1.0855 atm Temperature (T): 21°C = 21 + 273.15 = 294.15 K Volume (V): 145 L

Step 2: Calculate moles of H₂ gas

With the given conditions, we can now use the ideal gas law equation to determine the amount of \(\mathrm{H}_{2}\) produced (in moles): \( PV = nRT \) \( n = \frac{PV}{RT} \) \( n_{H_2} = \frac{(1.0855\,\text{atm})(145\,\text{L})}{(0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}})(294.15\,\text{K})} \) \( n_{H_2} = 6.8169 \,\text{moles} \)

Step 3: Stoichiometry of CaH₂

Now we determine the moles of \(\mathrm{CaH}_{2}\) required for generating moles of \(\mathrm{H}_{2}\) obtained above, according to the balanced chemical equation: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ As we can see, 1 mole of \(\mathrm{CaH}_{2}\) produces 2 moles of \(\mathrm{H}_{2}\). Therefore, we have the relationship: \( n_{CaH_2} = \frac{1}{2} n_{H_2} \)

Step 4: Calculate moles of CaH₂

By plugging the value of \(n_{H_2}\) we calculated earlier, we can find the moles of \(\mathrm{CaH}_{2}\): \( n_{CaH_2} = \frac{1}{2}(6.8169\,\text{moles}) \) \( n_{CaH_2} = 3.4084\,\text{moles} \)

Step 5: Calculate mass of CaH₂

Finally, we find the mass of required \(\mathrm{CaH}_{2}\) by multiplying the moles by its molar mass: Molar mass of \(\mathrm{CaH}_{2}\) = 40.08 (for Ca) + 2 * 1.01 (for H) = 42.10 g/mol Mass of \(\mathrm{CaH}_{2}\) = \( n_{CaH_2} \cdot \text{Molar mass of CaH}_{2} \) Mass of \(\mathrm{CaH}_{2}\) = \( 3.4084 \,\text{moles} \times 42.10 \,\frac{\text{g}}{\text{mol}} \) Mass of \(\mathrm{CaH}_{2}\) = 143.493 g Therefore, 143.493 g of \(\mathrm{CaH}_{2}\) are needed to generate 145 L of \(\mathrm{H}_{2}\) gas under the given conditions.

Key Concepts

Ideal Gas Law

Understanding the ideal gas law is essential for predicting the behavior of gases under various conditions. It's represented by the equation:
\[ PV = nRT \]
where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvins.

This law allows us to calculate one of the parameters if we know the others. In the context of the exercise, it was used to find the number of moles of hydrogen gas. Highlighting the importance of maintaining standard units, remember to convert the pressure into atmospheres (atm) and temperature to Kelvin (K), which facilitates the use of the ideal gas constant \( R = 0.0821 \frac{L\cdot atm}{mol\cdot K} \).
Applying the ideal gas law in practical scenarios, such as inflating life rafts or weather balloons, is an example of its real-world applications. In terms of education, it could be helpful to conduct experiments where students can visually see the effects of changing pressure, temperature, or volume on a gas, reinforcing the concept.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, called products. These reactions follow the law of conservation of mass, meaning the mass of the reactants equals the mass of the products. The reaction involving calcium hydride (\( \mathrm{CaH}_2 \)) and water is an example of a chemical reaction that releases hydrogen gas (
\( \mathrm{H}_2 \)). This type of reaction is also an example of a single-replacement reaction, where an element in a compound is replaced by another element.
When solving problems related to chemical reactions, it's crucial to start with a balanced chemical equation. The balanced equation for the reaction given in the exercise is:
\[ \mathrm{CaH}_{2}(s) + 2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(aq) + 2 \mathrm{H}_{2}(g) \]
From this, we can observe the stoichiometric relationship between calcium hydride and hydrogen gas, which states that 1 mole of \( \mathrm{CaH}_2 \) produces 2 moles of \( \mathrm{H}_2 \). This ratio is pivotal in calculating the amount of reactant needed for a desired amount of product. By understanding stoichiometry, one can predict the outcomes of reactions and calculate yields, essential for fields such as pharmaceuticals, manufacturing, and environmental science.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of that substance, which is the molecular weight measured in grams. It's a bridge between the macroscopic and microscopic worlds because it allows us to convert between the amount of substance in moles and its mass in grams. For the exercise, the calculation of molar mass is fundamental when we need to find out how many grams of calcium hydride (\( \mathrm{CaH}_2 \)) are required to produce a certain volume of hydrogen gas.
To calculate the molar mass, one adds up the atomic masses of all the atoms in the molecule. For \( \mathrm{CaH}_2 \), it includes the mass of one calcium atom and two hydrogen atoms:
\[ \text{Molar mass of } \mathrm{CaH}_{2} = 40.08\, (\text{for Ca}) + 2 \times 1.01\, (\text{for H}) = 42.10\, \frac{\text{g}}{\text{mol}} \]
With the molar mass, you can directly calculate the mass of the substance needed by multiplying the moles of substance by its molar mass. This process is showcased in the final step of the problem, allowing the conversion from the abstract amount of substance in moles to a tangible quantity of \( \mathrm{CaH}_2 \) in grams that can be measured and used in the laboratory or real-life applications.