Question

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react \(52.9 \mathrm{~g}\) of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide \(\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)\) would be produced? $$ \mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad \text { (unbalanced) } $$

Short answer

When 52.9 g of potassium chlorate reacts with excess red phosphorus, 20.4 g of tetraphosphorus decaoxide (P4O10) is produced.

Step-by-step solution

Balance the chemical equation

To balance the chemical equation, we need to make sure there are equal numbers of atoms of each element on both sides of the equation. The balanced equation is: $$ 6 \mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s)+6\mathrm{KCl}(s) $$

Finding moles of KClO3

First we need to find the moles of potassium chlorate (KClO3) given that we have 52.9 g of it. The molar mass of KClO3 is calculated as follows: Molar mass (KClO3) = 39.10(g/mol for K) + 35.45(g/mol for Cl) + 3 * 16(g/mol for O) = 122.55 g/mol Now, we can find the moles of KClO3: Moles(KClO3) = \(\frac{52.9 \mathrm{~g}}{122.55 \mathrm{~g/mol}}\) = 0.432 moles

Using stoichiometry to find moles of P4O10 produced

Now, we will use stoichiometry to find the moles of tetraphosphorus decaoxide (P4O10) produced. According to the balanced chemical equation, 6 moles of KClO3 react with 1 mole of P4O10. We can write this ratio as: Moles(P4O10) = Moles(KClO3) * \(\frac{1 \text{ mole } \mathrm{P}_{4}\mathrm{O}_{10}}{6 \text{ moles } \mathrm{KClO}_{3}}\) Moles(P4O10) = 0.432 moles * \(\frac{1}{6}\) = 0.072 mol

Calculate mass of P4O10 produced

Now, we will calculate the mass of P4O10 produced using its molar mass. The molar mass of P4O10 is calculated as follows: Molar mass (P4O10) = 4 * 30.97(g/mol for P) + 10 * 16(g/mol for O) = 283.88 g/mol To find the mass of P4O10 produced: Mass(P4O10) = Moles(P4O10) * Molar mass(P4O10) Mass(P4O10) = 0.072 mol * 283.88 g/mol = 20.4 g So, when 52.9 g of potassium chlorate reacts with excess red phosphorus, 20.4 g of tetraphosphorus decaoxide (P4O10) is produced.

Key concepts

Chemical Equation Balancing

In chemical reactions, balancing equations is crucial. It ensures the conservation of atoms and mass. Balancing a chemical equation means making sure that the number of each type of atom is the same on both sides of the equation. This step is vital to accurately predict the amounts of products and reactants involved in the reaction.
Let's look at the example of potassium chlorate reacting with red phosphorus. Initially, the chemical equation is unbalanced:- \(\mathrm{KClO}_{3}(s) + \mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) + \mathrm{KCl}(s)\)
The process of balancing requires adjusting the coefficients. Here, it results in:- \(6 \mathrm{KClO}_{3}(s) + \mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) + 6\mathrm{KCl}(s)\)
This balanced equation indicates that six molecules of potassium chlorate react with one molecule of phosphorus to produce one molecule of tetraphosphorus decaoxide and six molecules of potassium chloride. Always make sure to balance your equations—it is the foundation for every other calculation in stoichiometry.

Molar Mass Calculation

Molar mass calculations are essential to convert between mass and moles of a substance. The molar mass of a compound is the mass of one mole of its molecules, usually expressed in grams per mole (g/mol). In our exercise, determining the molar mass of potassium chlorate (KClO3) is an important step. The components and their atomic masses are: - K: 39.10 g/mol - Cl: 35.45 g/mol - O (x3): 16.00 g/mol each
Adding them up gives: - Molar mass of KClO3 = 39.10 + 35.45 + 3*16.00 = 122.55 g/mol.
Having the molar mass allows us to convert a given mass of KClO3 (like 52.9 g) into moles, which is crucial for stoichiometric calculations and to eventually find out how much product can be formed.

Reaction Stoichiometry

Reaction stoichiometry helps us understand the quantitative relationships between reactants and products in a chemical reaction. It utilizes the balanced equation to find out how much of each substance is involved or produced.In the balanced equation:- \(6 \mathrm{KClO}_{3}(s) + \mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) + 6 \mathrm{KCl}(s)\)
The ratio of potassium chlorate to tetraphosphorus decaoxide is 6:1. This means that for every 6 moles of KClO3, 1 mole of P4O10 is produced.
By knowing the amount of KClO3 in moles (0.432 moles from the given mass), we can use this stoichiometric ratio to find the amount of P4O10:- Moles of P4O10 = Moles of KClO3 x \(\frac{1}{6}\)Therefore, you can calculate that 0.431 moles of KClO3 yield 0.072 moles of P4O10.

Mass Calculation

Once you have calculated the moles of a product using stoichiometry, you can find the total mass. This is where the molar mass of the product comes into play again. For tetraphosphorus decaoxide (P4O10), first calculate its molar mass:

• P (x4): 30.97 g/mol each
• O (x10): 16.00 g/mol each

Add these to get: - Molar mass of P4O10: 4*30.97 + 10*16.00 = 283.88 g/mol
To find the mass of P4O10 produced, multiply the moles of P4O10 by its molar mass: - Mass = 0.072 moles * 283.88 g/mol Thus, the reaction produces 20.4 grams of P4O10.
Mass calculations are essential in determining yield and efficiency in chemical reactions.