Problem 41 - Chemistry The Central Science

Question

A first-stage recovery of magnesium from seawater is precipitation of \(\mathrm{Mg}(\mathrm{OH})_{2}\) with 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 lb of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\)

Step-by-Step Solution

Answer

To precipitate 1000 lb of Mg(OH)₂ from seawater, \(201,457.12 \ \text{grams}\) of CaO are needed.

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

Given 1000 lb of Mg(OH)₂, we will convert it to grams using the conversion factor: 1 lb = 453.592 g. So, the mass of Mg(OH)₂ in grams is: 1000 lb × 453.592 g/lb = 453,592 g

Step 2: Determine moles of Mg(OH)₂

Next, we will use the molar mass of Mg(OH)₂ to determine the number of moles. The molar mass of Mg(OH)₂ is 58.319 g/mol (Mg) + 2 × (15.999 g/mol (O) + 1.007 g/mol (H)) = 58.319 g/mol + 2 × 34.017 g/mol = 58.319 g/mol + 68.034 g/mol = 126.353 g/mol. Now, we can determine the moles: Moles of Mg(OH)₂ = (mass of Mg(OH)₂) / (molar mass of Mg(OH)₂) = 453,592 g / 126.353 g/mol ≈ 3591.68 mol

Step 3: Determine moles of CaO needed

From the balanced chemical equation, we can see that 1 mole of CaO is needed to precipitate 1 mole of Mg(OH)₂: Mg²⁺(aq) + CaO(s) + H₂O(l) → Mg(OH)₂(s) + Ca²⁺(aq) Hence, the moles of CaO needed are equal to the moles of Mg(OH)₂, which is 3591.68 mol.

Step 4: Convert moles of CaO to grams

Finally, we will convert the moles of CaO to grams, using its molar mass, which is 40.078 g/mol (Ca) + 15.999 g/mol (O) = 56.077 g/mol. Mass of CaO needed = (moles of CaO) × (molar mass of CaO) = 3591.68 mol × 56.077 g/mol ≈ 201,457.12 g Therefore, 201,457.12 grams of CaO are needed to precipitate 1000 lb of Mg(OH)₂.

Key Concepts

Chemical Reactions

A chemical reaction occurs when substances interact, leading to the formation of new products. This happens due to the rearrangement of atoms as chemical bonds are broken and new ones are formed. Chemical reactions are represented by chemical equations, which show the reactants on the left, an arrow in the middle, and the products on the right. In our example, we look at the reaction between magnesium ions from seawater and calcium oxide (CaO) to form magnesium hydroxide (Mg(OH)₂) and calcium ions.

• Reactants: These are the starting materials. Here they include magnesium ions (Mg²⁺), calcium oxide (CaO), and water (H₂O).
• Products: These are the substances formed. They include magnesium hydroxide (Mg(OH)₂) and calcium ions (Ca²⁺).
• Balanced Equation: To conserve mass, chemical equations must be balanced, meaning the number of each type of atom is the same on both sides. Here, 1 mole of CaO reacts with 1 mole of Mg²⁺ to yield 1 mole of Mg(OH)₂.

Understanding how to interpret this equation is crucial for stoichiometry, which is the calculation of reactants and products in chemical reactions.

Molar Mass

Molar mass is a fundamental concept in chemistry. It refers to the mass of one mole of a given substance, which is the molecular weight expressed in grams per mole. Calculating molar mass allows us to convert between grams and moles, a common task in chemistry.

• Calculation: To find the molar mass of a compound, sum the atomic masses of all the atoms present in its chemical formula. For instance, to find the molar mass of Mg(OH)₂, we calculate:

\[58.319 \text{ g/mol (Mg)} + 2 \times 34.017 \text{ g/mol (OH)} = 126.353 \text{ g/mol}\]

• Use in Reactions: Knowing the molar mass allows us to determine how much of each reactant is needed to completely react or how much product can be expected. Given that Mg(OH)₂ forms from Mg²⁺ and CaO in a 1:1 mol ratio, understanding its molar mass enables conversion from grams to moles and vice versa.

This underlines the importance of accurate molar mass calculations in predicting and measuring outcomes of reactions.

Unit Conversion

Unit conversion is vital in making sure that all measurements are consistently in the same units for accurate calculations. When tackling stoichiometry problems, it often involves converting between mass, moles, and sometimes volume.

• Mass to Grams: Often, weights are given in pounds (lbs), especially in countries using Imperial units. To engage in stoichiometric calculations, these must be converted into grams (SI units) using the conversion factor: 1 lb = 453.592 g.

In this example, the weight of Mg(OH)₂ was converted:
\[1000\, \text{lb} \times 453.592\, \text{g/lb} = 453,592\, \text{g}\]

• Other Common Conversions: Besides mass, converting volumes from liters to milliliters, or converting temperatures between Celsius and Kelvin, might be necessary depending on the problem context.

Unit conversion is an essential skill, simplifying communication between diverse scientific disciplines and ensuring clarity in results.