Problem 56 - Chemistry The Central Science In Si Units

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 \(110 \mathrm{kPa}\) at \(21^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

137.15 grams of \(\mathrm{CaH}_{2}\) are needed.

Step 1: Identify the Required Formula

To find the mass of calcium hydride needed, we will use the Ideal Gas Law to find the number of moles of hydrogen gas, and then use stoichiometry to find the mass of \(\mathrm{CaH}_{2}\) needed. The Ideal Gas Law is given by \(\mathbf{PV = nRT}\).

Step 2: Convert Given Values to Correct Units

Convert the temperature from Celsius to Kelvin by adding 273.15: \(21^{\circ} \mathrm{C} + 273.15 = 294.15\, \mathrm{K}\). Use the pressure in \(\mathrm{Pa}\) by converting \(110\,\mathrm{kPa}\) to \(110,000\,\mathrm{Pa}\).

Step 3: Calculate the Moles of Hydrogen Gas (n)

Using the Ideal Gas Law \[\mathbf{PV = nRT}\]: Set pressure \(P = 110,000\ \mathrm{Pa}\), volume \(V = 145 \, \mathrm{L} = 0.145 \, \mathrm{m^{3}}\), the gas constant \(R = 8.314 \, \mathrm{J/(mol\cdot K)}\), and temperature \(T = 294.15 \, \mathrm{K}\).Substitute the values into the equation:\[n = \frac{PV}{RT} = \frac{(110,000 \, \mathrm{Pa})(0.145 \, \mathrm{m^{3}})}{8.314 \, \mathrm{J/(mol\cdot K)} \times 294.15 \, \mathrm{K}}\]This gives \(n \approx 6.511\, \mathrm{mol}\) of hydrogen gas.

Step 4: Relate Moles of Hydrogen to Moles of \(\mathrm{CaH}_{2}\)

From the balanced chemical equation, each mole of \(\mathrm{CaH}_{2}\) yields 2 moles of \(\mathrm{H}_{2}\). Thus, moles of \(\mathrm{CaH}_{2}\) needed is \(= \frac{1}{2}\times6.511\) mol \(\approx 3.256\, \mathrm{mol}\).

Step 5: Calculate the Mass of \(\mathrm{CaH}_{2}\)

The molar mass of \(\mathrm{CaH}_{2}\) is calculated as follows: \(\mathrm{Ca}: 40.08\, \mathrm{g/mol}, \mathrm{H}: 2\times1.01\, \mathrm{g/mol}\), leading to \(\approx 42.10\, \mathrm{g/mol}\).Multiply the moles of \(\mathrm{CaH}_{2}\) by its molar mass to find the mass:\[3.256 \, \mathrm{mol} \times 42.10 \, \mathrm{g/mol} = 137.15 \, \mathrm{g}\]Thus, 137.15 grams of \(\mathrm{CaH}_{2}\) are needed.

Key Concepts

Stoichiometry

Stoichiometry plays a crucial role in chemistry as it helps us understand the relationships between the quantities of reactants and products in a chemical reaction. In this exercise, we are interested in determining how much calcium hydride (\(\mathrm{CaH}_{2}\)) is required to produce a specific volume of hydrogen gas (\(\mathrm{H}_{2}\)). To do this, we use stoichiometry principles to convert moles of \(\mathrm{H}_{2}\) gas, derived from the Ideal Gas Law, into moles of \(\mathrm{CaH}_{2}\).
Firstly, we start with the balanced chemical equation:

• \(\mathrm{CaH}_{2}(s) + 2 \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(aq)+2\mathrm{H}_{2}(g)\)

This equation tells us that one mole of \(\mathrm{CaH}_{2}\) produces two moles of \(\mathrm{H}_{2}\). Using this stoichiometric relationship, we can calculate the moles of \(\mathrm{CaH}_{2}\) needed by halving the moles of hydrogen gas needed. Thus, stoichiometry helps bridge the gap between the gaseous product and the solid reactant.
In line with the provided solution, the stoichiometrically required moles of \(\mathrm{CaH}_{2}\) were found to be approximately 3.256 moles. This exemplifies a fundamental stoichiometric calculation where the conversion factors are derived from the balanced chemical equation.

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to transform reactants into products. The key objective is understanding how substances interact under certain conditions. In our scenario, the focus is on the reaction between calcium hydride and water to produce calcium hydroxide and hydrogen gas.
The general skeleton of a chemical reaction can be represented as:

• Reactants \(\longrightarrow\) Products

In this case, our reactants are \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\), which react to form \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{H}_{2}\). The importance here lies in the ability of \(\mathrm{CaH}_{2}\) to act as a solid source of hydrogen gas.
Such chemical reactions are not just theoretical; they have practical applications. For example, the generation of hydrogen gas from the reaction can be used in inflating life rafts or other devices where a lightweight and portable source of gas is required. This application showcases the utility of chemistry in real-world scenarios and emphasizes the significance of understanding chemical reactions in both industrial and emergency settings.

Gas Laws

Gas laws help us understand and predict the behavior of gases under various conditions of pressure, volume, and temperature. The ideal gas law equation \(\mathbf{PV = nRT}\) is central to solving this exercise.
To utilize the Ideal Gas Law, we first need to ensure that all parameters are correctly converted to the appropriate units:

• Pressure \(P\) is converted from 110 \, \mathrm{kPa} to 110,000 \, \mathrm{Pa}.
• Volume \(V\) is used directly in liters but converted to cubic meters \(0.145 \, \mathrm{m^3}\) to ensure unit consistency.
• Temperature \(T\) must be in Kelvin, and we do this by adding 273.15 to the Celsius temperature.

Once we have these values, the formula can be rearranged to find the number of moles (\(n\)): \[n = \frac{PV}{RT}\]. Through this calculation, we determine that approximately \(6.511 \mathrm{mol}\) of \(\mathrm{H}_{2}\) are produced.
Understanding and correctly applying the gas laws allow us to bridge the gap between the theoretical and practical applications of chemistry, providing insights into how gases behave in different environments and influencing broader scientific applications, from weather forecasting to chemical manufacturing.