Question

Phosphorus can be prepared from calcium phosphate by the following reaction: \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) \longrightarrow\) \(6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g)\) Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other non-phosphorus- containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from \(1.0 \mathrm{~kg}\) of phosphorite if the phorphorite sample is \(75 \% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

Short answer

The maximum amount of P4 that can be produced from 1.0 kg of phosphorite containing 75% Ca3(PO4)2 by mass is approximately 150 g.

Step-by-step solution

1. Calculate the mass of Calcium Phosphate in the Phosphorite sample

To determine the mass of Ca3(PO4)2 in the phosphorite sample, we will multiply the total mass of the phosphorite sample (1.0 kg) by the percentage of Ca3(PO4)2 available in it (75%). Mass of Ca3(PO4)2 = (total mass of phosphorite sample) x (percentage of Ca3(PO4)2) Mass of Ca3(PO4)2 = (1.0 kg) x (0.75) Mass of Ca3(PO4)2 = 0.75 kg

2. Convert the mass of Calcium Phosphate to moles

Next, we need to find the number of moles of Ca3(PO4)2 present in the phosphorite sample. For this, we will use the molar mass of Ca3(PO4)2, which is 310.18 g/mol. First, we need to convert the mass of Ca3(PO4)2 from kg to grams: Mass of Ca3(PO4)2 = 0.75 kg x (1000 g/kg) = 750 g Now, we can calculate the number of moles: Number of moles of Ca3(PO4)2 = (mass of Ca3(PO4)2) / (molar mass of Ca3(PO4)2) Number of moles of Ca3(PO4)2 = (750 g)/(310.18 g/mol) ≈ 2.42 mol

3. Find the moles of P4 produced using stoichiometry

From the balanced equation, we can see that 2 moles of Ca3(PO4)2 produce 1 mole of P4. Therefore, we will use the stoichiometry of the reaction to find the number of moles of P4 produced. Number of moles of P4 produced = (Number of moles of Ca3(PO4)2) x (1 mole of P4 / 2 moles of Ca3(PO4)2) Number of moles of P4 produced = (2.42 mol) x (1/2) = 1.21 mol

4. Calculate the mass of P4 produced

Now that we know the number of moles of P4 produced, we can calculate the mass of P4. The molar mass of P4 is 123.88 g/mol. Mass of P4 produced = (number of moles of P4) x (molar mass of P4) Mass of P4 produced = (1.21 mol) x (123.88 g/mol) ≈ 150 g The maximum amount of P4 that can be produced from 1.0 kg of phosphorite is approximately 150 g.

Key concepts

Chemical Reactions

In the world of chemistry, chemical reactions are processes where substances, known as reactants, are transformed into new substances called products. These reactions involve the breaking and forming of chemical bonds and are fundamental to understanding material transformations.
The exercise presented involves a reaction for producing phosphorus from calcium phosphate, silica, and carbon.
Understanding this reaction helps us to see how phosphorus, a vital chemical element, is isolated from minerals. Chemical equations provide a symbolic representation of the reaction, indicating the reactants needed and the products formed.

• Chemical equations must be balanced, ensuring the same number of each element's atoms are present on both sides of the equation.
• The given equation shows a balanced reaction between calcium phosphate, silicon dioxide (silica), and carbon to produce phosphorus, calcium silicate, and carbon monoxide.

Phosphorus Production

Phosphorus is an essential element, widely used in fertilizers, animal feed, and detergents. The production of phosphorus from mineral ores involves several steps, the first of which is a chemical reduction reaction as seen in the exercise.
This reaction is not only a fundamental chemistry problem but a real-world industrial application. The process begins with extracting phosphorus from calcium phosphate, a major component of phosphate rocks, such as phosphorite.

• This synthesis requires silica and carbon in excess to ensure a complete transformation.
• Through this reaction, phosphorus is isolated while calcium silicate and carbon monoxide are formed as by-products.

The goal of the exercise is to find the maximum achievable output of phosphorus, assuming a specific concentration of calcium phosphate in the mineral.

Calcium Phosphate

Calcium phosphate is an inorganic compound that is a primary source of phosphorus in nature. It's found extensively in phosphate rocks and serves as a pivotal input in phosphorus production.
In the context of the problem, calcium phosphate appears in phosphorite, containing 75% of the mineral by mass.

• To calculate phosphorus production, it's crucial to first determine the mass of pure calcium phosphate present in the sample.
• This is done by multiplying the total mass of the phosphorite by 0.75, given its percentage composition.

Upon obtaining the mass, we convert it to moles using its molar mass—an essential step for stoichiometric calculations.

Molar Mass Calculation

Molar mass is a critical concept when dealing with chemical reactions and stoichiometry. It allows us to relate the mass of a substance to the amount in moles, essential for quantitative chemical analysis.
The molar mass of a compound is calculated by summing the atomic masses of all the atoms in its formula.

• For calcium phosphate ( Ca₃(PO₄)₂ ), the molar mass is calculated by taking into account the atomic masses of calcium (Ca), phosphorus (P), and oxygen (O).
• The total molar mass is an essential conversion factor for translating between lab-scale masses and chemical formula ratios.

For instance, calcium phosphate's molar mass is used to determine how many moles of this compound are present in a given mass. Hence, once we calculate the moles, stoichiometry guides us in determining the expected production of the target compound—in this case, phosphorus.