Question

For each of the following balanced reactions, calculate how many moles of product would be produced by complete conversion of 0.15 mol of the reactant indicated in boldface. State clearly the mole ratio used for the conversion. a. \(2 \mathbf{M} g(s)+O_{2}(g) \rightarrow 2 M g O(s)\) b. \(2 \mathrm{Mg}(s)+\mathbf{O}_{2}(g) \rightarrow 2 \mathrm{MgO}(s)\) c. 4 Fe \((s)+3 O_{2}(g) \rightarrow 2 F e_{2} O_{3}(s)\) d. \(4 \mathrm{Fe}(s)+3 \mathbf{O}_{2}(g) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\)

Short answer

Summary: a. 0.15 moles Mg -> 0.15 moles MgO using a mole ratio of 2 Mg : 2 MgO. b. 0.15 moles O2 -> 0.30 moles MgO using a mole ratio of 1 O2 : 2 MgO. c. 0.15 moles Fe -> 0.075 moles Fe2O3 using a mole ratio of 4 Fe : 2 Fe2O3. d. 0.15 moles O2 -> 0.10 moles Fe2O3 using a mole ratio of 3 O2 : 2 Fe2O3.

Step-by-step solution

Identify the given moles of reactant

In this reaction, we are given 0.15 moles of Mg (boldface). We will use this to find the moles of product, MgO.

Find the mole ratio

Examine the balanced equation to determine the mole ratio between Mg and MgO. From the equation, the ratio is: 2 moles of Mg : 2 moles of MgO

Calculate the moles of product

Using the mole ratio found in step 2, we calculate the moles of MgO produced: \(0.15\,\text{moles Mg} \cdot \frac{2\,\text{moles MgO}}{2\,\text{moles Mg}} = 0.15\,\text{moles MgO}\) #b. 2 Mg(s) + O2(g) -> 2 MgO(s)#

Identify the given moles of reactant

In this reaction, we are given 0.15 moles of O2 (boldface). We will use this to find the moles of product, MgO.

Find the mole ratio

Examine the balanced equation to determine the mole ratio between O2 and MgO. From the equation, the ratio is: 1 mole O2 : 2 moles of MgO

Calculate the moles of product

Using the mole ratio found in step 2, we calculate the moles of MgO produced: \(0.15\,\text{moles O}_2 \cdot \frac{2\,\text{moles MgO}}{1\,\text{mole O}_2} = 0.30\,\text{moles MgO}\) #c. 4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)#

Identify the given moles of reactant

In this reaction, we are given 0.15 moles of Fe (boldface). We will use this to find the moles of product, Fe2O3.

Find the mole ratio

Examine the balanced equation to determine the mole ratio between Fe and Fe2O3. From the equation, the ratio is: 4 moles of Fe : 2 moles of Fe2O3

Calculate the moles of product

Using the mole ratio found in step 2, we calculate the moles of Fe2O3 produced: \(0.15\,\text{moles Fe} \cdot \frac{2\,\text{moles Fe}_2\text{O}_3}{4\,\text{moles Fe}} = 0.075\,\text{moles Fe}_2\text{O}_3\) #d. 4 Fe(s) + 3 O2(g) -> 2 Fe2O3(s)#

Identify the given moles of reactant

In this reaction, we are given 0.15 moles of O2 (boldface). We will use this to find the moles of product, Fe2O3.

Find the mole ratio

Examine the balanced equation to determine the mole ratio between O2 and Fe2O3. From the equation, the ratio is: 3 moles of O2 : 2 moles of Fe2O3

Calculate the moles of product

Using the mole ratio found in step 2, we calculate the moles of Fe2O3 produced: \(0.15\,\text{moles O}_2 \cdot \frac{2\,\text{moles Fe}_2\text{O}_3}{3\,\text{moles O}_2} = 0.10\,\text{moles Fe}_2\text{O}_3\)

Key concepts

Stoichiometry

Imagine you're a chef, following a recipe to make a cake. You know that the ratio of eggs to flour is crucial; too much or too little of any ingredient can lead to disastrous results. Similarly, stoichiometry is the science of measuring the exact amounts of reactants and products involved in chemical reactions.

Just like following a cake recipe, stoichiometry involves ratios, but in this case, it's the mole ratio, which defines the proportion of substances reacting with each other. To predict how much product we get from a certain amount of reactant, we use these mole ratios, giving us a clear view of the chemical recipe. This becomes particularly useful when you encounter problems like calculating the number of moles of product generated by a given number of moles of reactant, as in the provided exercise.

Chemical Reactions

Chemical reactions are processes where one or more substances, the reactants, are transformed into new substances, the products. Understanding these chemical changes is much like observing a magic trick, where you see one thing turn into another.

In the magic of chemistry, the trick is called a balanced chemical equation. It shows us exactly what and how much we need for the reaction and what we'll get after. The key to solving stoichiometry problems is to first understand the chemical reaction—what's reacting and what's forming. The exercises provided demonstrate how to read these magical sentences and use them to find out precisely how much product forms from the reactants.

Molar Conversions

When cooking, you might need to convert cups to tablespoons. In chemistry, we do similar conversions, but here we convert units of measurement involving moles. This process is called molar conversion. It helps us switch between moles, mass, number of particles, volume of a gas, or even energy content, depending on what the reaction equation asks for.

The concept of 'mole' allows chemists to count atoms, molecules, or ions by weighing them. Molar conversion typically involves using Avogadro's number and molar mass as conversion factors. The exercise examples carefully show how to use the molar ratios from the balanced equations and convert moles of one substance to moles of another to find out how much product we can expect.

Balanced Chemical Equations

The foundation of working out any stoichiometry problem is a balanced chemical equation. This is just like balancing a scale; both sides need to be equal. In the context of chemical reactions, it means that the number of atoms for each element must be the same on both sides of the equation.

Understanding how to balance equations is vital as it is directly related to the mole ratios that are the cornerstone of stoichiometry. The sample exercises present balanced reactions and highlight their importance in determining the moles of reactants and products. By ensuring the conservation of mass, these balanced equations guide you through the precise calculations necessary for stoichiometric determinations.