Problem 54 - Chemistry Central Science

Question

Calcium hydride, \(\mathrm{CaH}_{2}\), reacts with water to form hydrogen gas: $$ \mathrm{CaH}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(I) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{2}(g) $$ This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating \(\mathrm{H}_{2}\) is desired. How many grams of \(\mathrm{CaH}_{2}\) are needed to generate \(53.5 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas if the pressure of \(\mathrm{H}_{2}\) is 814 torr at \(21{ }^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

52.2 grams of Calcium Hydride (CaH₂) are needed to generate 53.5 L of hydrogen gas under the given conditions of 814 torr pressure and 21°C temperature.

Step 1: Convert pressure to atm

The given pressure is in torr, but we need to use atm in the ideal gas law equation. To convert the pressure to atm, use the following conversion factor: \(1\ \mathrm{atm} = 760\ \mathrm{torr}\). So, $$ P = \frac{814\ \mathrm{torr}}{760\ \mathrm{torr/atm}} = 1.071\ \mathrm{atm} $$

Step 2: Convert temperature to Kelvin

We need to use the temperature in Kelvin when working with the ideal gas law equation. To convert the temperature to Kelvin, add 273.15 to the Celsius temperature: $$ T = 21°C + 273.15 = 294.15\ \mathrm{K} $$

Step 3: Use the ideal gas law to find moles of H₂ gas

Use the ideal gas law (PV = nRT) to find the moles of hydrogen gas needed. Rearranging the equation to solve for 'n', we have \(n = \frac{PV}{RT}\). Given values: P = 1.071 atm, V = 53.5 L, R = 0.0821 \( \dfrac{L \cdot atm}{mol \cdot K}\), T = 294.15 K. Now, plug in these values: $$ n(H_{2}) = \frac{(1.071\ \mathrm{atm})(53.5\ \mathrm{L})}{(0.0821\ \dfrac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(294.15\ \mathrm{K})} = 2.479\ \mathrm{moles\ of\ H_{2}} $$

Step 4: Use stoichiometry to find moles of CaH₂

From the balanced chemical equation, we can see that 1 mole of CaH₂ produces 2 moles of H₂ gas. So, we can set up a proportion to find the moles of CaH₂: $$ \frac{1\ \mathrm{mol\ of\ CaH_{2}}}{2\ \mathrm{mol\ of\ H_{2}}} = \frac{x\ \mathrm{mol\ of\ CaH_{2}}}{2.479\ \mathrm{mol\ of\ H_{2}}} $$ Solve for x (moles of CaH₂): $$ x = \frac{(1\ \mathrm{mol\ of\ CaH_{2}})(2.479\ \mathrm{mol\ of\ H_{2}})}{2\ \mathrm{mol\ of\ H_{2}}} = 1.2395\ \mathrm{moles\ of\ CaH_{2}} $$

Step 5: Convert moles of CaH₂ to grams

To find the grams of CaH₂ required, multiply the moles by the molar mass of CaH₂. The molar mass of CaH₂ is \(40.08\ \mathrm{g/mol} + (2 × 1.01\ \mathrm{g/mol}) = 42.10\ \mathrm{g/mol}\). $$ \text{grams of CaH}_{2} = (1.2395\ \mathrm{moles\ of\ CaH_{2}})(42.10\ \mathrm{g/mol}) = 52.2\ \mathrm{g\ of\ CaH_{2}} $$ So, 52.2 grams of Calcium Hydride (CaH₂) are needed to generate 53.5 L of hydrogen gas under the given conditions.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas. The law is typically written as PV = nRT, where R is the universal gas constant. For students dealing with gases, understanding this equation is essential since it allows the prediction of one property when the others are known.

The key to applying the ideal gas law to solve problems lies in first ensuring all units are consistent. Temperatures must be in Kelvin (K), volume in liters (L), pressure in atmospheres (atm), and the amount of gas in moles (mol). Converting units, such as pressure from torr to atm as demonstrated in the solution above, and temperatures from Celsius to Kelvin, is a critical step often missed by students. Remember, 0°C is equal to 273.15K, and 1 atm equals 760 torr. Keeping these conversion factors in mind will set a solid base for successfully using the Ideal Gas Law in stoichiometric calculations.

Stoichiometric Calculations

Stoichiometric calculations are at the heart of many chemical reactions in which reactants and products are quantitatively related. Through stoichiometry, we use the coefficients from balanced chemical equations to understand the relationships between reactants and products. It's like a recipe; if you know how much of one ingredient you have, you can figure out how much of the other ingredients you need. In the exercise, the chemical equation provided the ratio needed to convert moles of hydrogen gas to moles of calcium hydride.

To improve understanding, it's vital to grasp the concept of a 'mole ratio', which in this case tells us that 1 mole of \text{CaH}\(_2\) produces 2 moles of H\(_2\) gas. This ratio allows us to perform calculations to convert between moles of different substances. While the math might seem straightforward, students often struggle with the concept when the mole ratio isn't simple or when multiple reactants and products are involved. For a robust stoichiometric calculation, always start with a balanced equation and ensure that you are clear about the mole ratios involved before proceeding with mathematical conversions.

Molar Mass

Molar mass, a term frequently used in chemistry, is the mass of one mole of a substance (usually in grams per mole). It's an intrinsic property of each compound, based on the atomic masses of its constituent elements and how many of each appear in the formula. Molar mass enables us to convert between mass and moles, an essential step in most stoichiometric problems, such as the one from the exercise.

The molar mass of calcium hydride (CaH\(_2\)), for instance, is calculated by adding the molar mass of calcium to twice the molar mass of hydrogen, because the formula indicates there are two hydrogen atoms for every calcium atom. Understanding how to find and use molar mass is a critical skill for students; it's the bridge between the microscopic world of atoms and molecules and the macroscopic world we live in, where we measure substances in grams and liters. One area for improvement in solving these kinds of problems can be accuracy in the use of molar masses—students should ensure they are using the correct and precise atomic masses for elements when performing calculations.