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 phosphorus (P₄) that can be produced from 1.0 kg of phosphorite containing 75% calcium phosphate by mass is approximately 49.88 g.

Step-by-step solution

Calculate the mass of calcium phosphate

Since the given phosphorite sample contains 75% calcium phosphate by mass, the mass of calcium phosphate in the 1.0 kg sample can be calculated as follows: Mass of calcium phosphate = 75% of 1.0 kg = 0.75 kg.

Determine the number of moles of calcium phosphate

To determine the number of moles of calcium phosphate, divide the mass of calcium phosphate by its molar mass. The molar mass of Ca₃(PO₄)₂ is given by: Molar mass of Ca₃(PO₄)₂ = 3 × (mass of Ca) + 2 × (mass of P + 4 × mass of O). Plugging in the values from the periodic table: - Mass of Ca = 40.08 g/mol - Mass of P = 30.97 g/mol - Mass of O = 16.00 g/mol Molar mass of Ca₃(PO₄)₂ = 3 × 40.08 + 2 × (30.97 + 4 × 16.00) = 310.18(3) + 30.97(2) + 16.00(8) = 930.54 g/mol Now, calculate the number of moles of calcium phosphate: Moles of Ca₃(PO₄)₂ = \(\frac{0.75\ kg}{930.54\ \frac{g}{mol}}\) = \(\frac{750\ g}{930.54\ \frac{g}{mol}}\) = 0.8061 mol (4 s.f.)

Use stoichiometry to determine the number of moles of phosphorus (P₄) that can be produced

The balanced chemical equation is: 2 Ca₃(PO₄)₂ + 6 SiO₂ + 10 C → 6 CaSiO₃ + P₄ + 10 CO From this equation, we can see that 2 moles of calcium phosphate yield 1 mole of phosphorus (P₄). Therefore, the number of moles of phosphorus (P₄) that can be produced is given by: Moles of P₄ = \(\frac{1}{2}\) × moles of Ca₃(PO₄)₂ = \(\frac{1}{2}\) × 0.8061 mol = 0.40305 mol (5 s.f.)

Calculate the mass of phosphorus that can be produced

To find the mass of phosphorus that can be produced, multiply the number of moles of P₄ by its molar mass. The molar mass of P₄ is given by: Molar mass of P₄ = 4 × mass of P = 4 × 30.97 g/mol = 123.88 g/mol Now, calculate the mass of phosphorus that can be produced: Mass of P₄ = moles of P₄ × molar mass of P₄ = 0.40305 mol × 123.88 g/mol = 49.88 g (4 s.f.) Therefore, the maximum amount of phosphorus (P₄) that can be produced from 1.0 kg of phosphorite containing 75% calcium phosphate by mass is approximately 49.88 g.

Key concepts

Mass-Mole Conversion

When dealing with chemical stoichiometry, understanding mass-mole conversion is crucial. This process involves converting the mass of a substance to moles using its molar mass. For instance, in the original exercise, you start with 1.0 kg of phosphorite, which is 75% calcium phosphate (Ca₃(PO₄)₂).

The first step is to calculate the actual mass of calcium phosphate in the sample: 75% of 1.0 kg equals 0.75 kg. Next, to find the number of moles of calcium phosphate, use its molar mass. The molar mass calculation for Ca₃(PO₄)₂ involves adding together the atomic masses of calcium, phosphorus, and oxygen, resulting in approximately 310.54 g/mol.

Once you have the molar mass, divide the mass of calcium phosphate by this molar mass to determine the moles of Ca₃(PO₄)₂. This calculation (0.75 kg = 750 g, divided by 310.54 g/mol) yields approximately 0.8061 moles of calcium phosphate. This conversion is fundamental because it allows chemists to relate macroscopic masses to the number of particles involved in reactions.

Limiting Reactant

The concept of a limiting reactant is central to predicting the amount of product formed in a chemical reaction. This reactant is entirely consumed first and limits the extent of the reaction. In the given problem, despite other reactants like SiO₂ and C being in excess, calcium phosphate (Ca₃(PO₄)₂) is the focus.

The balanced equation provided is critical for identifying the stoichiometric relationship between reactants and products: - 2 Ca₃(PO₄)₂ react with an excess of 6 SiO₂ and 10 C - Produces 1 mole of P₄ for every 2 moles of Ca₃(PO₄)₂

This tells us that the amount of Ca₃(PO₄)₂ dictates how much P₄ can be formed. By knowing that 0.8061 moles of Ca₃(PO₄)₂ are available, you can use stoichiometry to determine the moles of P₄ produced. Since it takes 2 moles of Ca₃(PO₄)₂ to produce 1 mole of P₄, the limiting reactant (Ca₃(PO₄)₂) ultimately determines the maximum yield of P₄.

Chemical Equations

Chemical equations are a concise way to describe chemical reactions using symbols and formulas for the reactants and products. They must be balanced to obey the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system.

In the exercise, the balanced chemical equation helps us understand the stoichiometric relationships:\(2 \text{ Ca}_3(\text{PO}_4)_2 + 6 \text{ SiO}_2 + 10 \text{ C} \rightarrow 6 \text{ CaSiO}_3 + \text{ P}_4 + 10 \text{ CO}\)

This equation shows:- For every 2 moles of Ca₃(PO₄)₂, you need 6 moles of SiO₂ and 10 moles of C.- This produces 1 mole of P₄ and 6 moles of CaSiO₃, plus 10 moles of CO.

Balancing this equation ensures that the number of atoms of each element is the same on both sides of the equation. It ensures accuracy in predicting the quantities of reactants and products involved.

Mole-Mass Calculations

Mole-mass calculations are essential for converting between the mass of reactants or products and the number of moles. This is a recurrent method used in stoichiometry to progress from moles of a substance calculated through a mass-mole conversion to the mass of a different substance in the reaction.

In this exercise, after calculating that 0.40305 moles of P₄ can be produced, the next step is to find the mass of phosphorus. This involves using the molar mass of P₄, which is 123.88 g/mol. Multiply the moles of P₄ by this molar mass to find the mass:\(\text{Mass of } \text{P}_4 = 0.40305 \text{ mol} \times 123.88 \text{ g/mol} = 49.88 \text{ g}\)

Such calculations allow chemists to relate the theoretical yields (like the 49.88 g of P₄) with practical scenarios in lab settings. Understanding these calculations makes it easier to scale reactions and predict amounts of products from given reactants.