Question

Phosphorus can be prepared from calcium phosphate by the following reaction: $$\begin{aligned}2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+& 10 \mathrm{C}(s) \longrightarrow \\\& 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) \end{aligned}$$ 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 is approximately \(149.79 \, \text{g}\).

Step-by-step solution

Calculate the mass of calcium phosphate in the phosphorite sample

Since the phosphorite sample is 75% calcium phosphate by mass, we can calculate the mass of calcium phosphate in the 1.0 kg phosphorite sample as follows: Mass of calcium phosphate = (1.0 kg phosphorite) × (0.75) = 0.75 kg Since we will be dealing with grams throughout the problem, let's convert the mass to grams: 0.75 kg = 750 g

Determine the moles of calcium phosphate

To determine the moles of calcium phosphate, we need to find the molar mass of Ca₃(PO₄)₂: Molar mass of Ca₃(PO₄)₂ = (3 × molar mass of Ca) + (2 × molar mass of P) + (8 × molar mass of O) = (3 × 40.08 g/mol) + (2 × 30.97 g/mol) + (8 × 16.00 g/mol) = 310.18 g/mol Now, let's calculate the moles of calcium phosphate in the sample: Moles of Ca₃(PO₄)₂ = (750 g Ca₃(PO₄)₂) / (310.18 g/mol) = 2.418 mol

Use stoichiometry to calculate the moles of P₄ produced

The balanced chemical equation gives us the stoichiometric ratio between the moles of Ca₃(PO₄)₂ and moles of P₄ produced: 2 moles of Ca₃(PO₄)₂ → 1 mole of P₄ Using this ratio, we can calculate the moles of P₄ produced from the moles of Ca₃(PO₄)₂: Moles of P₄ = (1 mol P₄ / 2 mol Ca₃(PO₄)₂) × (2.418 mol Ca₃(PO₄)₂) = 1.209 mol P₄

Convert the moles of P₄ to grams

To find the mass of P₄ produced, we need to find the molar mass of P₄: Molar mass of P₄ = 4 × molar mass of P = 4 × 30.97 g/mol = 123.88 g/mol Now, let's calculate the grams of P₄ produced: Mass of P₄ = (1.209 mol P₄) × (123.88 g/mol) = 149.79 g The maximum amount of P₄ that can be produced from 1.0 kg of phosphorite is approximately 149.79 g.

Key concepts

Chemical Reactions

Chemical reactions are a fascinating process where substances, called reactants, interact to form new substances, known as products. In this particular reaction, phosphorus is prepared from calcium phosphate, silicon dioxide, and carbon. The reaction can be summarized with the chemical equation:- \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} + 6 \mathrm{SiO}_{2} + 10 \mathrm{C} \rightarrow 6 \mathrm{CaSiO}_{3} + \mathrm{P}_{4} + 10 \mathrm{CO}\)This equation shows us that the reactants are calcium phosphate (\(\mathrm{Ca}_3(\mathrm{PO}_4)_2\)), silicon dioxide (\(\mathrm{SiO}_2\)), and carbon (\(\mathrm{C}\)), which combine to form calcium silicate (\(\mathrm{CaSiO}_3\)), phosphorus (\(\mathrm{P}_4\)), and carbon monoxide (\(\mathrm{CO}\)).
Chemical reactions often require a balancing process to reflect the Law of Conservation of Mass, ensuring the same number of each type of atom exists on both sides of the equation.

Molar Mass

Molar mass is an important concept in understanding chemical reactions and stoichiometry. It refers to the mass of one mole of a given substance, usually expressed in grams per mole (g/mol).
To compute the molar mass, you add up the atomic masses of all atoms in a molecule.- For calcium phosphate (\(\mathrm{Ca}_3(\mathrm{PO}_4)_2\)), calculate as follows: - 3 calcium atoms: \(3 \times 40.08\) g/mol - 2 phosphorus atoms: \(2 \times 30.97\) g/mol - 8 oxygen atoms: \(8 \times 16.00\) g/molThus, the molar mass of \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\) is calculated as: \(310.18\) g/mol.
Molar mass allows chemists to convert between the mass of a substance and the amount in moles, which is crucial for using stoichiometry in balancing chemical reactions and determining the amounts of products formed.

Balanced Chemical Equations

Balanced chemical equations are crucial for accurately describing a chemical reaction. They ensure that the amount of each element is conserved, abiding by the Law of Conservation of Mass.
To balance a chemical equation, adjust the coefficients (the numbers before the molecules) so that the number of atoms for each element is the same on both sides of the equation.
In our example, the equation:- \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} + 6 \mathrm{SiO}_{2} + 10 \mathrm{C} \rightarrow 6 \mathrm{CaSiO}_{3} + \mathrm{P}_{4} + 10 \mathrm{CO}\)shows balance as:- 6 calcium and 4 phosphorus atoms on both sides- 24 oxygen atoms and 20 carbon atoms on both sidesBy balancing chemical equations, we can use stoichiometry to predict the amounts of substances consumed and produced in a reaction. This is a crucial tool for solving complex quantitative chemistry problems.