Question

A laboratory method of preparing ${\mathrm{O}}_2(\mathrm{g})$ involves the decomposition of ${\mathrm{KClO}}_3(\mathrm{s})$ $$2{\mathrm{KClO}}_3(\mathrm{s})\stackrel{\mathrm{\Delta }}{⟶}2\mathrm{KCl}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})$$ (a) How many moles of ${\mathrm{O}}_2(\mathrm{g})$ can be produced by the decomposition of $32.8\mathrm{g}{\mathrm{KClO}}_3?$ (b) How many grams of ${\mathrm{KClO}}_3$ must decompose to produce $50.0\mathrm{g}{\mathrm{O}}_2?$ (c) How many grams of KCl are formed, together with $28.3\mathrm{g}{\mathrm{O}}_2,$ in the decomposition of ${\mathrm{KClO}}_3?$

Short answer

The answers are: (a) 0.402 moles of ${\mathrm{O}}_2$ can be produced from 32.8 g of ${\mathrm{KClO}}_3$; (b) 127.4 g of ${\mathrm{KClO}}_3$ must be decomposed to produce 50.0 g of ${\mathrm{O}}_2$; (c) 43.9 g of $\mathrm{KCl}$ are formed along with 28.3 g of ${\mathrm{O}}_2$.

Step-by-step solution

Converting given grams to moles (for question a)

Calculate the amount of moles of ${\mathrm{KClO}}_3$ from the given mass using its molar mass, which can be calculated as $39.10+35.45+(3\times 16)=122.55\mathrm{g}/\mathrm{mol}$.\n So, the number of moles is $32.8\mathrm{g}/122.55\mathrm{g}/\mathrm{mol}=0.268\mathrm{mol}$.

Using mole-ratio to calculate moles of product

Use the stoichiometric ratio between ${\mathrm{KClO}}_3$ and ${\mathrm{O}}_2$ (2:3) to calculate the moles of ${\mathrm{O}}_2$ formed. This gives $0.268\mathrm{mol}\times (3/2)=0.402\mathrm{mol}{\mathrm{O}}_2$.

Converting moles to grams to find required reactant mass (for question b)

Convert the given mass of ${\mathrm{O}}_2$ into moles using its molar mass. The molar mass of ${\mathrm{O}}_2$ is $32.00\mathrm{g}/\mathrm{mol}$, so its moles are $50.0\mathrm{g}/32.00\mathrm{g}/\mathrm{mol}=1.56\mathrm{mol}$. Then use the stoichiometric ratio between ${\mathrm{KClO}}_3$ and ${\mathrm{O}}_2$ (2:3) to find the moles of ${\mathrm{KClO}}_3$ decomposed, which gives $1.56\mathrm{mol}\times (2/3)=1.04\mathrm{mol}$. The mass of ${\mathrm{KClO}}_3$ is then calculated as $1.04\mathrm{mol}\times 122.55\mathrm{g}/\mathrm{mol}=127.4\mathrm{g}$.

Calculating mass of product with known mass of another product (for question c)

Convert the mass of ${\mathrm{O}}_2$ to moles. The moles of ${\mathrm{O}}_2$ are $28.3\mathrm{g}/32.00\mathrm{g}/\mathrm{mol}=0.884\mathrm{mol}$. Use the stoichiometric ratio between ${\mathrm{O}}_2$ and $\mathrm{KCl}$ (3:2) to find the moles of $\mathrm{KCl}$ formed, which gives $0.884\mathrm{mol}\times (2/3)=0.589\mathrm{mol}$. The molar mass of $\mathrm{KCl}$ is $74.55\mathrm{g}/\mathrm{mol}$, so the mass of $\mathrm{KCl}$ is $0.589\mathrm{mol}\times 74.55\mathrm{g}/\mathrm{mol}=43.9\mathrm{g}$.

Key concepts

Mole Calculations

Understanding mole calculations is crucial when working with chemical reactions because it allows you to interconvert mass and number of molecules efficiently. The mole is a fundamental unit in chemistry that represents a specific number of molecules or atoms, namely Avogadro's number, which is approximately $6.022\times {10}^{23}$. When you have a mass of a substance, like $32.8\text{g of }{\mathrm{KClO}}_3$, you first need to calculate the moles by using the molar mass of the substance.

For instance, the molar mass of ${\mathrm{KClO}}_3$ is determined by adding the atomic masses of its constituent elements: Potassium (K), Chlorine (Cl), and Oxygen (O). The calculation results in $122.55\text{g/mol}$. To find the moles of ${\mathrm{KClO}}_3$ in $32.8\text{g}$ of the compound, divide the mass by the molar mass: $32.8\text{g}/122.55\text{g/mol}=0.268\text{mol}$. This conversion is important because chemical equations and stoichiometry are based on moles, not mass.

Chemical Decomposition

Chemical decomposition is a vital reaction type where a compound breaks down into simpler substances; in this case, ${\mathrm{KClO}}_3$ decomposes into $\mathrm{KCl}$ and ${\mathrm{O}}_2$. This process is often facilitated by heating, indicated by the delta symbol $\mathrm{\Delta }$ over the reaction arrow. Decomposition reactions are generally endothermic, meaning they absorb energy rather than release it.

The reaction can be represented as: $$2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2$$
This shows the breakdown of two moles of ${\mathrm{KClO}}_3$ into two moles of $\mathrm{KCl}$ and three moles of oxygen gas ${\mathrm{O}}_2$. Understanding this chemical breakdown helps us predict the products formed in a reaction, and highlights why mole calculations are necessary to determine specific quantities for substances involved.

Oxygen Production

In the decomposition of ${\mathrm{KClO}}_3$, one of the key products is oxygen gas ${\mathrm{O}}_2$. Calculating the amount of oxygen produced involves understanding how to use stoichiometric ratios effectively. This reaction is an excellent example of how substances decompose to yield products necessary for various applications, such as biological respiration or chemical processes.

To determine how much ${\mathrm{O}}_2$ is generated, we use its molar mass, $32.00\text{g/mol}$, and the stoichiometry from the reaction equation. For example, starting with $32.8\text{g of }{\mathrm{KClO}}_3$, we found that it results in $0.402\text{mol}{\mathrm{O}}_2$. If needed in mass terms, converting moles to grams gives the weight of the produced ${\mathrm{O}}_2$ by multiplying the moles by its molar mass.
By understanding how much oxygen is produced from a given amount of ${\mathrm{KClO}}_3$, we can gauge how much such reactions contribute to requirements in industrial or environmental settings.

Stoichiometric Ratios

Stoichiometric ratios are the heart of chemical stoichiometry, and they are determined from the coefficients of a balanced chemical equation. These ratios allow us to calculate the amounts of reactants required to produce a specific amount of products, and vice versa.

In the given decomposition reaction: $$2{\mathrm{KClO}}_3\to 2\mathrm{KCl}+3{\mathrm{O}}_2$$
the stoichiometric ratio between ${\mathrm{KClO}}_3$ and ${\mathrm{O}}_2$ is 2:3. This means for every 2 moles of ${\mathrm{KClO}}_3$, 3 moles of ${\mathrm{O}}_2$ are produced. Such ratios are used to convert between moles of different substances within a chemical equation.

• For question (a), with 0.268 mol ${\mathrm{KClO}}_3$, using the ratio gives $0.402\text{mol}{\mathrm{O}}_2$.
• For question (b), starting with $50.0\text{g of }{\mathrm{O}}_2$, you first calculate the moles of oxygen, and then use the ratio to find the needed moles (and hence mass) of ${\mathrm{KClO}}_3$ to produce that amount.

Understanding these relationships ensures precise predictions in chemical processes, highlighting the power of stoichiometric principles in practical applications.