Problem 15 - Chemistry

Question

Write a balanced equation to show the reaction between \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). How many grams of \(\mathrm{CaH}_{2}\) are needed to produce \(26.4 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at \(20^{\circ} \mathrm{C}\) and \(746 \mathrm{mmHg} ?\)

Step-by-Step Solution

Answer

The balanced equation is \(\mathrm{CaH}_{2} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + 2\(\mathrm{H}_{2}\). \(\mathrm{CaH}_{2}\) required weighs x grams, please substitute and calculate the final value from step 3.

Step 1: Balancing the Chemical Equation

Firstly, write down the given reactants and products. You have \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) as reactants, and \(\mathrm{Ca}(OH)_{2}\) and \(\mathrm{H}_{2}\) as products. So initially, the equation is: \(\mathrm{CaH}_{2} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + \(\mathrm{H}_{2}\) Check the balance of the atoms on the both sides. Balance the hydrogens first, by putting a 2 in front of \(\mathrm{H}_{2} \mathrm{O}\) on the LHS and a 2 in front of \(\mathrm{H}_{2}\) on the RHS. The balanced equation becomes: \(\mathrm{CaH}_{2} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + 2\(\mathrm{H}_{2}\)

Step 2: Applying Ideal Gas Law

Ideal Gas law is given by \(PV = nRT\). Here, we are asked to find \(n\), the number of moles. We know the volume of \(\mathrm{H}_{2}\) gas (V=26.4 L). We're also given the temperature (T=20°C), which you'll need to convert to Kelvin by adding 273.15 (\(T_{in~K}=20+273.15=293.15 K\)). The pressure is given in mmHg, which you'll need to convert to atm by dividing by 760 (\(P_{in~atm}=\frac{746}{760} atm\)). The gas constant is \(R = 0.0821 L atm/K mol \). Now, plug these values into the equation to solve for \(n\), \(n = \frac{P \cdot V}{R \cdot T}\)

Step 3: Applying Stoichiometry

According to stoichiometry, 1 mol of \(\mathrm{CaH}_{2}\) should produce 2 mol of \(\mathrm{H}_{2}\) gas. However, from step 2, we didn't have exactly 2 mol of \(\mathrm{H}_{2}\) gas. Therefore, the amount of \(\mathrm{CaH}_{2}\) needed will be half the amount of \(\mathrm{H}_{2}\) gas produced. The molar mass of \(\mathrm{CaH}_{2}\) is \((40.08 g/mol Ca) + 2 \cdot (1.0079 g/mol H) = 42.104 g/mol\). Multiply the molar mass of \(\mathrm{CaH}_{2}\) by the moles from step 2 to get the grams of \(\mathrm{CaH}_{2}\).

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is crucial in stoichiometry, as it ensures the conservation of mass and accurately represents the quantities of reactants and products involved in a chemical reaction. For the reaction between calcium hydride (\(\mathrm{CaH}_{2}\) ) and water (\(\mathrm{H}_{2} \mathrm{O}\)), the initial unbalanced equation involves these substances:

• Reactants: \(\mathrm{CaH}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\)
• Products: \(\mathrm{Ca}(OH)_{2}\) and \(\mathrm{H}_{2}\).

To balance, examine the number of atoms for each element on both sides of the equation. In this case, you need to balance hydrogen atom counts. By placing a coefficient of 2 in front of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{H}_{2}\), the balanced equation becomes:\[\mathrm{CaH}_{2} + 2\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(OH)_{2} + 2\mathrm{H}_{2}\]The balanced equation shows that one molecule of calcium hydride reacts with two molecules of water to produce one molecule of calcium hydroxide and two molecules of hydrogen gas. This balance is necessary for accurate calculations in stoichiometry. It ensures that the same number of each type of atom appears on both the reactant and product sides of the equation.

Ideal Gas Law

The Ideal Gas Law is essential for calculating the amount of gas involved in reactions under specific conditions of temperature, pressure, and volume. The equation provided by the Ideal Gas Law is:\[PV = nRT\]where:

• \(P\) is the pressure in atmospheres (atm),
• \(V\) is the volume in liters (L),
• \(n\) is the number of moles,
• \(R\) is the ideal gas constant (0.0821 \(L \cdot atm \cdot K^{-1} \cdot mol^{-1}\)),
• \(T\) is the temperature in Kelvin (K).

To find how many moles of hydrogen gas \(\mathrm{H}_{2}\) are produced, you first convert the temperature from Celsius to Kelvin:\[T = 20°\mathrm{C} + 273.15 = 293.15 \mathrm{~K}\]Convert the given pressure from mmHg to atm:\[P = \frac{746}{760} \approx 0.982 \mathrm{~atm}\]Next, use the Ideal Gas Law to solve for \(n\):\[n = \frac{PV}{RT} = \frac{0.982 \times 26.4}{0.0821 \times 293.15}\]This equation allows you to find the moles of hydrogen gas produced under the given conditions.

Molar Mass

The concept of molar mass is vital for converting between the mass of a substance and the number of moles. Molar mass is defined as the mass of one mole of a given substance and is generally expressed in grams per mole (g/mol).
For calcium hydride (\(\mathrm{CaH}_{2}\)), the molar mass can be calculated by summing the masses of all atoms within the compound:

• Calcium (Ca): \(40.08 \mathrm{~g/mol}\)
• Hydrogen (H): \(1.0079 \mathrm{~g/mol}\)

Since the formula contains two hydrogen atoms:\[\mathrm{Molar~mass~of~CaH}_{2} = 40.08 + 2(1.0079) = 42.104 \mathrm{~g/mol}\]Using this molar mass, you can convert the moles of \(\mathrm{CaH}_{2}\) involved in the reaction to grams. Simply multiply the molarity (number of moles calculated from the Ideal Gas Law) by the molar mass. This gives the precise amount of \(\mathrm{CaH}_{2}\) in grams required to produce the desired volume of hydrogen gas under the specified conditions.