Question

How many moles of \(\mathrm{P}_{4}\) can be produced by reaction of \(0.10\) moles \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}, 0.36\) moles \(\mathrm{SiO}_{2}\) and \(0.90\) moles \(\mathrm{C}\) according to the following reaction ? $$ 4 \mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}+18 \mathrm{SiO}_{2}+30 \mathrm{C} \longrightarrow 3 \mathrm{P}_{4}+2 \mathrm{CaF}_{2}+18 \mathrm{CaSiO}_{3}+30 \mathrm{CO} $$ (a) \(0.060\) (b) \(0.030\) (c) \(0.045\) (d) \(0.075\)

Short answer

0.030 moles of \(\mathrm{P}_{4}\) can be produced.

Step-by-step solution

Write down the balanced chemical equation

First, ensure that the chemical equation is balanced and reflect the stoichiometry of the reactants and products involved in the reaction.

Calculate the mole ratio

From the balanced equation, find the mole ratio between the reactants and the product of interest, \(\mathrm{P}_{4}\).

Determine the limiting reactant

Compare the moles of each reactant to their required mole ratios to find out which reactant will be exhausted first, limiting the amount of product that can be produced.

Calculate moles of \(\mathrm{P}_{4}\) produced

Using the mole ratio and the quantity of the limiting reactant, calculate the number of moles of \(\mathrm{P}_{4}\) that can be formed.

Choose the correct answer

Match the calculated number of moles of \(\mathrm{P}_{4}\) with the given options to find the correct answer.

Key concepts

Limiting Reactant Calculation

When mixing different substances to initiate a chemical reaction, the limiting reactant is the one that will be completely used up first, thus determining the maximum amount of product that can be formed. Understanding how to calculate the limiting reactant is crucial for predicting the outcome of chemical reactions.

To identify the limiting reactant, one must first know the starting quantities of all reactants. In a stoichiometric problem, you compare the mole ratio of each reactant to the amount actually provided. The reactant that yields the least amount of product, when its available moles are divided by its stoichiometric coefficient, is the limiting reactant. For instance, if we have reactants A and B, entering into a reaction in a 1:3 ratio, and we start with an equal number of moles of both, A would be the limiting reactant, as we would need three times more of B for the reaction to proceed completely.

Mole Ratio Determination

Mole ratio determination is an integral part of stoichiometry, serving as the bridge that allows chemists to convert between amounts of different substances in a reaction. The mole ratio is derived from a balanced chemical equation and expresses the relationship between the reactants and products.

For example, in the balanced equation \( aA + bB \rightarrow cC + dD \), the mole ratio of \( A \rightarrow C \) is \( a:c \), implying that for every \( a \) moles of A consumed, \( c \) moles of C are produced. Determining mole ratios requires a balanced equation to ensure that the law of conservation of mass is respected. This step is indispensable for accurately calculating the amount of product formed from given quantities of reactants.

Balanced Chemical Equations

A balanced chemical equation epitomizes the principle of conservation of mass, indicating that matter is neither created nor destroyed in a chemical reaction. It ensures that the number of atoms for each element is the same on both the reactant and product sides of the equation.

To balance an equation, one adjusts the coefficients, which are the numbers placed before the chemical formulas. Balanced equations are essential in stoichiometry because they allow the mole ratio of reactants to products to be determined accurately. These ratios are used to predict the amounts of products that will form from given reactants. Without a balanced equation, any further computations, including determining the limiting reactant or calculating theoretical yields, would lead to incorrect results. The balance of the given reaction is crucial for solving stoichiometric problems.

JEE Chemistry Preparation

Competitive exams like the Joint Entrance Examination (JEE) in India are highly demanding, and Chemistry is one of the core subjects that require rigorous preparation. Students aiming to perform well on the exam must develop a strong understanding of concepts like stoichiometry, which includes calculations of limiting reactants, mole ratios, and balancing chemical equations.

Preparing for such topics goes beyond memorizing formulas; it requires practice and the application of concepts to solve complex problems. Practice problems from textbooks and previous exams, and review step-by-step solutions to enhance understanding. Utilizing online educational platforms to bridge gaps in knowledge can also be beneficial. Regularly engaging with these materials helps to build competence in solving stoichiometry problems, which are common in the JEE Chemistry section. Timed practice sessions can also help emulate exam conditions and improve time management skills for the actual test.