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 nonphosphorus- 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, which is 75% Ca3(PO4)2 by mass, is 149.78g.

Step-by-step solution

Calculate the moles of the limiting reactant Ca3(PO4)2 in the sample.

In this case, the limiting reactant is Ca3(PO4)2, as it is the only reactant affected by the mass of the phosphorite. We are given that the mineral is 75% Ca3(PO4)2 by mass, so we first find the mass of Ca3(PO4)2 in the 1 kg sample: Mass of Ca3(PO4)2 = (75/100) * 1000g = 750g Next, determine the molar mass of Ca3(PO4)2: Molar mass of Ca3(PO4)2 = (3 * 40.08g/mol Ca) + (2 * (1 * (30.97g/mol P) + 4 * (16g/mol O))) = 310.18g/mol Now, we can convert the mass of Ca3(PO4)2 to moles: Moles of Ca3(PO4)2 = 750g / 310.18g/mol = 2.418 mol

Use stoichiometry to determine the moles of P4 produced.

Based on the balanced chemical equation, we know that 2 moles of Ca3(PO4)2 produce 1 mole of P4. So we can set up a proportion to calculate the number of moles of P4 produced from the moles of Ca3(PO4)2: (1 mol P4 / 2 mol Ca3(PO4)2) = (x moles P4 / 2.418 mol Ca3(PO4)2) Cross-multiplying and solving for x, we get: x moles P4 = (1 mol P4) * (2.418 mol Ca3(PO4)2) / 2 mol Ca3(PO4)2 x moles P4 = 1.209 mol

Calculate the mass of P4 produced from the moles of P4.

Since we now have the moles of P4 produced, we can calculate the mass of P4 using its molar mass: Molar mass of P4 = 4 * (30.97g/mol P) = 123.88g/mol So, mass of P4 produced = 1.209 mol * 123.88g/mol = 149.78g Hence, the maximum amount of P4 that can be produced from 1.0 kg of phosphorite is 149.78g.

Key concepts

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It helps us understand how much reactant is needed to form a desired amount of product. In this case, stoichiometry lets us calculate the amount of phosphorus ( P_4 ) produced from a specific mass of phosphorite.

To tackle stoichiometric calculations, follow these steps:

• Write and balance the chemical equation. This provides the mole ratios for the substances involved.
• Convert the given quantities (mass, volume) to moles using molar masses or gas laws.
• Use mole ratios from the balanced equation to calculate moles of the desired compound.
• Convert the moles calculated back to desired units (mass, volume).

Understanding these steps can help you solve a wide range of problems in chemistry where amount relationships are crucial.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. In the phosphorus production reaction, calcium phosphate reacts with silicon dioxide and carbon to form calcium silicate, phosphorus, and carbon monoxide.

This transformation is an example of a simultaneous occurrence of reduction and oxidation, often termed a redox reaction. The reaction clearly shows a balanced equation, which is essential to correctly determine the stoichiometric coefficients. These coefficients are pivotal as they guide which quantities of reactants convert to products.

• Recognition of types of chemical reactions (e.g., synthesis, decomposition, redox) deepens understanding.
• Balancing chemical reactions ensures the law of conservation of mass is maintained.

Knowing the type of reaction and being able to balance it are foundational skills in mastering physical chemistry.

Limiting Reactant

Identifying the limiting reactant in a chemical reaction is crucial, as it determines the maximum amount of product that can be formed. In our example, calcium phosphate (Ca_3(PO_4)_2) is the limiting reactant because it's the only one whose quantity is initially limited by the composition of the phosphorite.

To find the limiting reactant:

• Calculate the moles of each reactant present.
• Use the balanced chemical equation to determine how much product each reactant can produce.
• The reactant that produces the least amount of product is the limiting reactant.

Knowing the limiting reactant ensures you maximize efficiency and minimize waste in chemical processes.

Mole Calculations

Mole calculations form the backbone of quantitative chemistry. They allow chemists to express the number of molecules or atoms in a given mass of a substance. From the phosphorite problem, we calculate the number of moles of necessary compounds to determine the maximum phosphorus production.

Here's a quick guide to mole calculations:

• Determine the molar mass of the substance by adding up the atomic masses of its elements.
• Convert the mass of a substance to moles using the relationship: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
• Repeat this process for each reactant or product in the equation if necessary.

Mastery of mole calculations is essential not just for phosphorus production, but for any reaction involving substances with known molar masses.