Problem 39 - Chemistry The Central Science

Question

A first-stage recovery of magnesium from seawater is precipitation of \(\mathrm{Mg}(\mathrm{OH})_{2}\) with \(\mathrm{CaO}\) : \(\mathrm{Mg}^{2+}(a q)+\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Ca}^{2+}(a q)\) What mass of \(\mathrm{CaO}\), in grams, is needed to precipitate \(1000 \mathrm{lb}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\)

Step-by-Step Solution

Answer

The mass of CaO required to precipitate 1000 lb of Mg(OH)₂ is approximately 436236.8 grams.

Step 1: Convert the mass of Mg(OH)₂ to grams

First, we are given the mass of Mg(OH)₂ in pounds (1000 lb), but we need to convert it to grams since our final answer requires grams. We know that: 1 lb = 453.592 g We'll multiply the given mass by the conversion factor: \( 1000 \, \text{lb} \times \frac{453.592 \, \text{g}}{1 \, \text{lb}} \)

Step 2: Calculate the moles of Mg(OH)₂

To perform stoichiometry and relate the given mass of Mg(OH)₂ to the mass of CaO, we first need to find the number of moles of Mg(OH)₂. The molar mass of Mg(OH)₂ = 24.305 g/mol (Mg) + 2(15.999 g/mol (O) + 1.008 g/mol (H)) = 58.319 g/mol Next, calculate the moles of Mg(OH)₂: \( \frac{1000 \times 453.592 \, \text{g}}{1 \, \text{lb}} \times \frac{1 \, \text{mol}}{58.319 \, \text{g}} \)

Step 3: Applying stoichiometry to find moles of CaO

According to the balanced chemical equation, 1 mole of CaO is required to precipitate 1 mole of Mg(OH)₂: \( \text{Mg}^{2+}(aq) + \text{CaO}(s) + \text{H}_{2} \text{O}(l) \longrightarrow \text{Mg}(\text{OH})_{2}(s) + \text{Ca}^{2+}(aq) \) So, the moles of CaO required is equal to the moles of Mg(OH)₂. With this information, we can now determine the mass of CaO in grams.

Step 4: Converting moles of CaO to mass in grams

To find the mass of CaO in grams, we'll need the molar mass of CaO. The molar mass of CaO = 40.08 g/mol (Ca) + 15.999 g/mol (O) = 56.079 g/mol. Now, let's multiply the moles of CaO by its molar mass: \( \frac{1000 \times 453.592 \, \text{g}}{1 \, \text{lb}} \times \frac{1 \, \text{mol}}{58.319 \, \text{g}} \times \frac{56.079 \, \text{g}}{1 \, \text{mol}} \)

Step 5: Solve for mass of CaO in grams

After calculating, we have: \( 1000 \times 453.592 \, \text{g} \times \frac{1 \, \text{mol}}{58.319 \, \text{g}} \times \frac{56.079 \, \text{g}}{1 \, \text{mol}} ≈ 436236.8 \, \text{g} \) Therefore, 436236.8 grams of CaO are needed to precipitate 1000 lb of Mg(OH)₂.

Key Concepts

Chemical Reactions

A chemical reaction occurs when substances, known as reactants, transform into new substances called products through a process of breaking and forming chemical bonds. The equation representing the reaction showcases the reactants on the left and the products on the right, with an arrow indicating the direction of the transformation.

In our exercise, we have a chemical reaction involving magnesium ions from seawater reacting with calcium oxide (CaO) and water to produce magnesium hydroxide (Mg(OH)₂) and aqueous calcium ions. The balanced chemical equation reflects a one-to-one stoichiometry between magnesium ions and calcium oxide; for each mole of CaO, one mole of Mg(OH)₂ is formed.

Understanding the stoichiometry of the reaction allows for the calculation of how much CaO is needed to completely react with a certain amount of Mg(OH)₂. The balanced equation is fundamental in ensuring that the conservation of mass is respected and all atoms accounted for in the process, which is also an expression of the Law of Conservation of Mass.

Molar Mass

Molar mass is a property of a substance defined as the mass of one mole of its particles (atoms, molecules, or formula units) and is expressed in gram per mole (g/mol). It's a bridge between the mass of a sample and the number of moles it contains.

For instance, as calculated in our exercise, the molar mass of magnesium hydroxide, Mg(OH)₂, is approximately 58.319 g/mol, determined by summing the molar masses of individual elements: magnesium, oxygen, and hydrogen. This calculation involves multiplying the atomic mass of each element by the number of atoms of that element in the formula and adding these values together. The same principle applies to calculate CaO's molar mass, ending up with 56.079 g/mol.

Knowing the molar mass is crucial for converting between the mass of a substance and the number of moles. This is important because moles are used in stoichiometry calculations to predict the amounts of reactants and products involved in a chemical reaction.

Moles to Grams Conversion

The conversion from moles to grams and vice versa is a fundamental aspect of stoichiometry in chemistry. It involves using the molar mass of a substance as a conversion factor. To convert from moles to grams, you multiply the number of moles by the molar mass of the substance.

In the context of our exercise, after determining the molar mass of both Mg(OH)₂ and CaO, we first convert the given mass of Mg(OH)₂ in pounds to grams using a unit conversion factor (1 lb = 453.592 g). Then, we use the molar mass of Mg(OH)₂ to find the number of moles. With the balanced chemical equation, we can conclude that the moles of CaO needed are equal to the moles of Mg(OH)₂.

The final step is to convert the moles of CaO back to grams, for which we use CaO's molar mass as the conversion factor. The result tells us the precise mass of CaO required to precipitate 1000 pounds of Mg(OH)₂, completing our stoichiometry calculation and providing the answer to the problem posed.