Problem 41 - Chemistry The Central Science In Si Units

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 \(1000 \mathrm{~kg}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\)

Step-by-Step Solution

Answer

962,000 grams of \(\mathrm{CaO}\) are needed.

Step 1: Understanding the Reaction

The chemical reaction involved in the precipitation of \(\mathrm{Mg(OH)}_2\) from magnesium ions using lime (\(\mathrm{CaO}\)) is given as \(\mathrm{Mg}^{2+}(aq) + \mathrm{CaO}(s) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{Mg(OH)}_2(s) + \mathrm{Ca}^{2+}(aq)\).

Step 2: Determine Molar Masses

Calculate the molar masses of the compounds involved: \(\mathrm{Mg(OH)}_2\) has a molar mass of 58.32 g/mol (\(\mathrm{Mg} = 24.31\, \mathrm{g/mol}, \mathrm{O} = 16.00\, \mathrm{g/mol}, \mathrm{H} = 1.01\, \mathrm{g/mol}\)). \(\mathrm{CaO}\) has a molar mass of 56.08 g/mol (\(\mathrm{Ca} = 40.08 \mathrm{g/mol}, \mathrm{O} = 16.00 \mathrm{g/mol}\)).

Step 3: Convert Mass to Moles of \(\mathrm{Mg(OH)}_2\)

Convert the given mass of \(\mathrm{Mg(OH)}_2\) to moles using its molar mass: \[\text{moles of } \mathrm{Mg(OH)}_2 = \frac{1000\,\mathrm{kg} \times 1000\,\mathrm{g/kg}}{58.32\,\mathrm{g/mol}}\]\[\text{moles of } \mathrm{Mg(OH)}_2 = 17150.41\,\text{moles}\]

Step 4: Determine Moles of \(\mathrm{CaO}\) Needed

The balanced chemical equation shows a 1:1 molar ratio between \(\mathrm{Mg(OH)}_2\) and \(\mathrm{CaO}\). Therefore, the moles of \(\mathrm{CaO}\) needed is also 17150.41 moles.

Step 5: Convert Moles of \(\mathrm{CaO}\) to Mass

Convert the moles of \(\mathrm{CaO}\) to mass using its molar mass: \[\text{mass of } \mathrm{CaO} = 17150.41\,\text{moles} \times 56.08\,\mathrm{g/mol}\]\[\text{mass of } \mathrm{CaO} = 961,998.97\,\mathrm{g}\]

Step 6: Finalize the Answer

Since the mass required must be presented in grams, the result is approximately 962,000 grams of \(\mathrm{CaO}\).

Key Concepts

Molar Mass Calculation

Calculating molar mass is a vital step in solving many chemistry problems. The molar mass is the mass of one mole of a substance and is measured in grams per mole (g/mol). Each element has a specific atomic mass, and by adding these atomic masses together, we find the molar mass of a compound.

For example, to find the molar mass of \(\mathrm{Mg(OH)}_2\), you need to sum the molar masses of magnesium (Mg), oxygen (O), and hydrogen (H). Magnesium has an atomic mass of 24.31 g/mol, oxygen is 16.00 g/mol, and hydrogen is 1.01 g/mol. Therefore:

\[\text{molar mass of } \mathrm{Mg(OH)}_2 = 24.31 + (16.00 \times 2) + (1.01 \times 2) = 58.32 \text{ g/mol} \]

Similarly, for calcium oxide (\(\mathrm{CaO}\)), the calculation would be:

\[\text{molar mass of } \mathrm{CaO} = 40.08 \text{ (Ca)} + 16.00 \text{ (O)} = 56.08 \text{ g/mol} \]

Understanding this calculation is crucial as it facilitates the conversion of mass to moles, which is often necessary for stoichiometry and chemical reactions.

Balanced Chemical Equations

A balanced chemical equation ensures that the same number of each type of atom appears on both the reactants and the products side of the equation. This balance reflects the conservation of mass, meaning that matter is neither created nor destroyed in a chemical reaction.

In our given equation: \(\mathrm{Mg}^{2+}(aq) + \mathrm{CaO}(s) + \mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{Mg(OH)}_2(s) + \mathrm{Ca}^{2+}(aq)\), the equation is already balanced because it has:

• 1 Magnesium (Mg) atom on both sides.


• 2 Oxygen (O) atoms both in the consuming \(\mathrm{CaO}\) and produced \(\mathrm{Mg(OH)}_2\).


• 1 Calcium (Ca) atom on both sides.


• And overall, a balance of ionic charge.

Understanding how to balance an equation is foundational in chemistry. It allows chemists to precisely compute the amount of reactants required or products formed in reactions.

Conversion of Units

Understanding and performing unit conversions is crucial in science. It allows you to transition between different measurement systems or units to ensure that all components are compatible for calculations.

In this example, we needed to convert kilograms to grams for the magnesium hydroxide mass. The conversion is straightforward: \(1\) kg is equivalent to \(1000\) grams. Therefore, \(1000\) kg equals \(1,000,000\) grams.

Different reactions might require conversions such as:

• Grams to moles, as performed using the molar mass.


• Liters to milliliters for volume measurements (\(1\) liter = \(1000\) milliliters).


• Celsius to Kelvin for temperature adjustments (\(\text{K} = \text{C} + 273.15\)).

Unit conversion enables precise calculation and is a transferable skill across various STEM fields.