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

Short answer

The mass of calcium hydride needed to generate 145 L of hydrogen gas at 110 kPa and 21°C is: \(m_{CaH_2} = \frac{1}{2} (\frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\:K)}) \times 42.10 \: g/mol \approx 926.35\: g\) Thus, approximately 926.35 g of CaH₂ is needed to generate 145 L of H₂ gas under the given conditions.

Step-by-step solution

Convert the volume, temperature, and pressure of hydrogen gas to moles

Given the volume (V) of hydrogen gas = 145 L, pressure (P) = 110 kPa, and temperature (T) = 21°C. We need to convert the temperature to Kelvin (K) by adding 273.15 to the Celsius temperature: T(K) = 21 + 273.15 = 294.15 K. Now, we'll use the Ideal Gas Law: \(PV = nRT\) Where - P is the pressure in kPa - V is the volume in L - n is the number of moles - R is the ideal gas constant [8.314 J/(mol·K) or 0.08206 L·kPa/(mol·K)] - T is the temperature in K Rearrange the Ideal Gas Law to solve for the number of moles (n) of hydrogen gas (H₂): \(n = \frac{PV}{RT}\). Now, plug in the known values: P = 110 kPa, V = 145 L, R = 0.08206 L·kPa/(mol·K), and T = 294.15 K: \(n_{H2} = \frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\: K)}\)

Use the stoichiometry of the balanced chemical equation to convert moles of hydrogen gas to moles of calcium hydride

According to the balanced chemical equation: 1 mole of CaH₂ (s) → 2 moles of H₂ (g) Now, use the stoichiometry to convert the moles of H₂ to moles of CaH₂: Let \(n_{CaH_2}\) be the number of moles of calcium hydride needed. \(\frac{n_{CaH_2}}{n_{H2}} = \frac{1}{2}\) \(n_{CaH_2} = \frac{1}{2} n_{H2}\) Now, calculate \(n_{CaH_2}\) using the value of \(n_{H2}\) obtained in the previous step: \(n_{CaH_2} = \frac{1}{2} (\frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\:K)})\)

Convert the moles of calcium hydride to grams using its molar mass

To calculate the mass of calcium hydride needed, we need to know its molar mass, which is given by: Molar mass of CaH₂ = Molar mass of calcium (Ca) + 2 × Molar mass of hydrogen (H) Molar mass of CaH₂ = 40.08 g/mol (Ca) + 2 × 1.01 g/mol (H) = 42.10 g/mol Now, apply the formula mass = moles × molar mass: Mass of CaH₂ = \(n_{CaH_2} \times\) molar mass of CaH₂ \(m_{CaH_2} = \frac{1}{2} (\frac{(110\: kPa)(145\: L)}{(0.08206\: L \cdot kPa/(mol \cdot K))(294.15\:K)}) \times 42.10 \: g/mol\) Calculate the mass of calcium hydride needed to generate 145 L of hydrogen gas at the given pressure and temperature.

Key concepts

Ideal Gas Law

Understanding the Ideal Gas Law is crucial when dealing with gases in chemical reactions. The Ideal Gas Law is a handy equation used to relate pressure, volume, temperature, and the number of moles of a gas. This law is expressed as:\[PV = nRT\]Here:

• P is the pressure of the gas, measured in kilopascals (kPa).
• V is the volume of the gas in liters (L).
• n represents the moles of gas.
• R is the ideal gas constant, valued at 0.08206 L·kPa/(mol·K) when pressure is in kPa.
• T is the temperature in Kelvin (K), easily obtained by adding 273.15 to the Celsius temperature.

To find out how many moles of hydrogen gas are produced, rearrange the equation to solve for \(n\):\[n = \frac{PV}{RT}\]For our specific reaction, plug in the measurements: pressure of 110 kPa, volume of 145 L, and a temperature of 294.15 K. With the Ideal Gas Law, you can calculate the moles of hydrogen necessary for a balanced reaction. Breaking the process into these steps helps make complex calculations easier to handle.

Chemical Reactions

Chemical reactions involve the transformation of substances into new products, commonly described using balanced chemical equations. In our example, we examine the reaction between calcium hydride (\(\text{CaH}_2\)) and water (\(\text{H}_2\text{O}\)), producing calcium hydroxide (\(\text{Ca(OH)}_2\)) and hydrogen gas (\(\text{H}_2\)).The balanced reaction is:\[\text{CaH}_2(s) + 2\,\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + 2\,\text{H}_2(g)\]Notice how this reaction equation is balanced, meaning the same number of each type of atom is present on both sides of the equation. Correct balancing ensures that the law of conservation of mass is obeyed, as matter is neither created nor destroyed.In every stoichiometric calculation, following the exact mole ratios provided in a balanced chemical equation is essential. This allows for proper prediction of the amounts of products generated from any given reactant.

Moles Calculation

Moles are a fundamental concept in chemistry, acting as a universal measure for counting atoms, molecules, or ions in a given sample. Calculating the moles involved in a reaction starts with the Ideal Gas Law, which provides a way to relate volume, pressure, and temperature of a gas to moles.Looking at the step where we need to determine the moles of calcium hydride (\(\text{CaH}_2\)): When hydrogen gas forms, the stoichiometry of the reaction shows that 1 mole of calcium hydride produces 2 moles of hydrogen gas.Given a certain amount of hydrogen gas, our calculations involve:

• Using the Ideal Gas Law to find moles of hydrogen (\(n_{H2}\)).
• Applying stoichiometry to find moles of calcium hydride, using the ratio \(\frac{1}{2}n_{H2}\).

To find the mass of \(\text{CaH}_2\) needed, use its molar mass, which is 42.10 g/mol:\[ \text{Mass} = n_{\text{CaH}_2} \times 42.10 \, \text{g/mol} \]These calculations show how the concepts of moles and stoichiometry help translate theoretical equations into practical product amounts in any given chemical reaction.