Problem 104 - Chemistry The Central Science

Question

A mixture containing \(\mathrm{KClO}_{3}, \mathrm{K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\) and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{array}{l}{2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)} \\ {2 \mathrm{KHCO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)} \\\ {\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)}\end{array} $$ The KCl does not react under the conditions of the reaction. If 100.0 g of the mixture produces 1.80 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and 4.00 \(\mathrm{g}\) of \(\mathrm{O}_{2},\) what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

Step-by-Step Solution

Answer

The composition of the original 100g mixture was: - KClO3: 10.21 g - KHCO3: 10.01 g - K2CO3: 13.82 g - KCl: 65.96 g

Step 1: Identify the stoichiometric coefficients and molar masses

From the given balanced chemical equations, we can identify the stoichiometric coefficients (the numbers in front of compounds that balance the equations) for each reaction. We also need the molar mass (grams/mole) of each component. Here are the values: - KClO3: \( 2 KClO_3 \rightarrow 2 KCl + 3 O_2\); Molar mass: 39.1 (K) + 35.5 (Cl) + 3*16 (O) = 122.6 g/mol - KHCO3: \( 2 KHCO_3 \rightarrow K_2 O + H_2 O + 2 CO_2\); Molar mass: 39.1 (K) + 1.01 (H) + 12.01 (C) + 3*16 (O) = 100.12 g/mol - K2CO3: \( K_2 CO_3 \rightarrow K_2 O + CO_2\); Molar mass: 2*39.1 (K) + 12.01 (C) + 3*16 (O) = 138.21 g/mol

Step 2: Determine moles of products formed

We need to determine the moles of each product (O2, H2O, and CO2) formed during the reaction. This is done by dividing the mass of each product by their respective molar masses: H2O: Mass: 1.80g; Molar mass: 18.02 g/mol; Moles: 1.80 / 18.02 = 0.1 mol O2: Mass: 4.00g; Molar mass: 2*16 = 32 g/mol; Moles: 4.00 / 32 = 0.125 mol CO2: Mass: 13.2g; Molar mass: 12.01 (C) + 2*16 (O) = 44.01 g/mol; Moles: 13.2 / 44.01 = 0.3 mol

Step 3: Calculate moles and mass of each reactant

Using the stoichiometric coefficients and the moles of the products formed, we can calculate the moles of each reactant, and then find their mass: 1. KClO3: Moles of O2 produced: 0.125 mol According to the balanced equation, 2 KClO3 -> 3 O2, so (0.125 * 2) / 3 = 0.0833 mol of KClO3. Mass = (0.0833 mol) * (122.6 g/mol) = 10.21 g 2. KHCO3: Moles of H2O and CO2 produced: 0.1 mol and 0.3 mol According to the balanced equation, 2 KHCO3 -> H2O + 2 CO2, so 0.1 mol H2O and (0.3/2) mol CO2 corresponds to 0.1 mol KHCO3. Mass = (0.1 mol) * (100.12 g/mol) = 10.01 g 3. K2CO3: Moles of CO2 produced: 0.3 mol CO2 from KHCO3 reaction: 0.1*2=0.2 mol (from Step 3) CO2 from K2CO3: 0.1 mol (0.3-0.2 mol) According to the balanced equation, K2CO3 -> CO2, so 0.1 mol CO2 corresponds to 0.1 mol K2CO3. Mass = (0.1 mol) * (138.21 g/mol) = 13.82 g 4. KCl: It doesn't react with other reactants and doesn't form any product. To find the mass of KCl, we will subtract the mass of the other three components from the initial mass of the mixture. Mass of KCl: Initial mass of mixture - (Mass of KClO3 + Mass of KHCO3 + Mass of K2CO3) Mass of KCl: 100g - (10.21g + 10.01g + 13.82g) = 65.96g The composition of the original mixture was: - KClO3: 10.21 g - KHCO3: 10.01 g - K2CO3: 13.82 g - KCl: 65.96 g

Key Concepts

Chemical Reactions

Chemical reactions transform substances into new products by rearranging their atoms. In the given exercise, different compounds containing potassium ( K), chloride ( Cl), and oxygen ( O) underwent transformations when heated. This resulted in the release of gases like CO2, O2, and H2O. To understand these reactions, it is crucial to be familiar with the concept of reactants and products as well as energy changes.

• Reactants are the starting materials that undergo a chemical change.
• Products are the new substances formed as a result of that change.
• Energy, often in the form of heat, is required to break bonds in reactants, and energy is released when new bonds form in products.

Analyzing these transformations involves understanding how different compounds decompose or combine, often requiring specific conditions like heating to trigger the reaction. In this problem, heating the compounds resulted in complete decomposition, which is a typical scenario in stoichiometry tasks to ensure all reactants are transformed into products.

Mole Calculations

The mole is a fundamental unit in chemistry, used to quantify the amount of a substance. In stoichiometry, mole calculations are essential as they allow chemists to relate mass to the number of particles or atoms. In the context of the exercise, calculating the moles of compounds produced from the mix was critical.

Each mole of a substance has a mass equal to its molar mass expressed in grams. The exercise demonstrates how to determine the moles from the mass of each gaseous product:

• To find moles, divide the given mass by the molar mass.
• For instance, from the exercise, 13.20 g of CO2 translates to 0.3 moles.

Understanding mole ratios is also important, which involves using coefficients from balanced equations to relate reactants and products. By determining the moles of products (like H2O and O2) formed, we can then back-calculate the moles of original compounds present, which helps in figuring out the composition of the starting mixture. Performing these mole calculations accurately is foundational in stoichiometric analysis, ensuring that all atoms are accounted for in chemical reactions.

Balanced Equations

Balanced equations are representations of chemical reactions where the number of atoms for each element is the same on both sides of the equation. This concept is rooted in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction, only rearranged.
In this exercise, equations like \[2 \mathrm{KClO}_{3} \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O}_{2}\] are balanced to ensure that every atom's count on the left equals the atom count on the right.

When interpreting and manipulating balanced equations, consider the following:

• Stoichiometric coefficients (the numbers in front of chemical formulas) indicate proportions of reactants and products involved.
• These coefficients are crucial for calculating how much of a substance is produced or consumed in a reaction.
• The correct balance reflects the actual nature of chemical processes, epitomizing how molecules react in fixed ratios.

Using these balanced equations, we calculated the mass of reactants needed to produce specific amounts of product. This problem-solving skill ensures chemists can predict product yields and reactant requirements accurately in real experiments, making it invaluable in laboratory and industrial settings.