Problem 396 - Chemistry: The Central Science

Question

A first-stage recovery of magnesium from seawater is precipitation of \(\mathrm{Mg}(\mathrm{OH})_{2}\) with \(\mathrm{CaO}\) : $$ \mathrm{Mg}^{2+}(a q)+\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Ca}^{2+}(a q) $$ 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

To precipitate 1000 lbs of magnesium hydroxide, 435988.90 grams of calcium oxide are needed.

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

To convert the given mass of magnesium hydroxide (1000 lbs) to moles, we need its molar mass. The molar mass of Mg(OH)2 can be calculated as: Molar mass of Mg(OH)2 \(=1 \times \mathrm{Mg} + 2 \times(\mathrm{H}+\mathrm{O})=24.31+2(1.01+16)=24.31+34.02=58.33 \mathrm{g/mol}\) Now we have to convert the mass of 1000 lbs to grams. Since 1 lb = 453.59 g, the mass is: \(1000 \mathrm{lb}=1000\times453.59\mathrm{g}= 453590\mathrm{g}\) To find the moles of Mg(OH)2: \(moles\,of\,Mg(OH)_2 = \frac{mass}{molar\,mass} = \frac{453590\,\mathrm{g}}{58.33\,\mathrm{g/mol}} = 7777.88\,\mathrm{mol}\)

Step 2: Use stoichiometry to find the moles of CaO needed

From the balanced chemical equation, we can see that 1 mole of Mg(OH)2 requires 1 mole of CaO. Therefore, the moles of calcium oxide needed will be equal to the moles of magnesium hydroxide that we have calculated. Moles of CaO \(= 7777.88\,\mathrm{mol}\)

Step 3: Convert moles of CaO to grams

Now, we need to convert the moles of CaO to grams. The molar mass of CaO can be calculated as: Molar mass of CaO \(=1 \times \mathrm{Ca} + 1 \times \mathrm{O} = 40.08 + 16 = 56.08\, \mathrm{g/mol}\) To find the mass of CaO needed: Mass of CaO \(= moles \times molar\,mass = 7777.88\,\mathrm{mol} \times 56.08\, \mathrm{g/mol} = 435988.90\,\mathrm{g}\) Therefore, 435988.90 grams of calcium oxide are needed to precipitate 1000 lbs of magnesium hydroxide.

Key Concepts

Molar Mass Calculation

Understanding the molar mass of a compound is a fundamental aspect of stoichiometry in chemical reactions. The molar mass is the weight of one mole of a substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms present in the compound.

For instance, the molar mass of magnesium hydroxide \( \mathrm{Mg}(\mathrm{OH})_{2} \) is calculated by adding the atomic mass of magnesium (Mg), which is 24.31 g/mol, to twice the sum of the atomic masses of hydrogen (H) and oxygen (O), because the compound includes two hydroxide groups. This results in a molar mass of 58.33 g/mol. Knowing this value allows for conversions between mass and moles, which is pivotal in quantitative chemistry.

Chemical Equation Balancing

Balancing chemical equations is a critical skill for solving stoichiometry problems. It involves making sure that the number of atoms for each element is equal on both sides of the equation. This reflects the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction.

In our exercise, the balanced chemical equation \( \mathrm{Mg}^{2+}(aq) + \mathrm{CaO}(s) + \mathrm{H}_{2}O(t) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) + \mathrm{Ca}^{2+}(aq) \) shows a one-to-one ratio between magnesium ions and calcium oxide. Such proper stoichiometric relationships are essential when determining the equivalent amounts of reactants and products involved.

Moles to Mass Conversion

In chemistry, converting between moles and mass is a routine calculation described by the equation, mass = moles \(\times\) molar mass. After determining the molar mass, it is used to convert the mass of a substance to the amount in moles and vice versa.

For our example, with a known molar mass of calcium oxide (CaO) being 56.08 g/mol, we can calculate the mass required to react with a given amount of magnesium hydroxide using the moles to mass conversion formula. This step is crucial in practical chemical preparations and experimental work, ensuring the correct proportions of substances are used.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble product, known as a precipitate. Such reactions are commonly used in various industrial processes, including the recovery of metals from solutions.

In the exercise, the precipitation of magnesium hydroxide from seawater involves reacting magnesium ions with calcium oxide to form the insoluble \(\mathrm{Mg}(\mathrm{OH})_{2}\). Understanding these types of reactions is not only important for executing them in the laboratory but also for applications such as water treatment and mineral recovery. The ability to predict and calculate the outcome of precipitation reactions is a valuable skill in chemistry.