Consider the generic chemical reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 3 \mathrm{C} $$ How many moles of B are required to completely react with: (a) \(6 \mathrm{~mol}\) of \(\mathrm{A}\) (b) \(2 \mathrm{~mol}\) of \(\mathrm{A}\) (c) \(7 \mathrm{~mol}\) of \(\mathrm{A}\) (d) 11 mol of \(\mathrm{A}\)

Short Answer

Expert verified
For complete reactions, (a) 9 mol, (b) 3 mol, (c) 10.5 mol, (d) 16.5 mol of B are required.

Step by step solution

01

Understanding the Stoichiometry

The stoichiometry of the reaction given by the equation is 2 moles of A react with 3 moles of B to produce 3 moles of C. To find out how many moles of B are required, we need to use the molar ratio between A and B.
02

Calculating Moles of B for 6 mol of A

Using the stoichiometric coefficients from the chemical equation, we set up a proportion. For every 2 moles of A, 3 moles of B are required. Therefore, for 6 moles of A: \[\begin{equation} \frac{6\,\text{mol A}}{2\,\text{mol A}} \times 3\,\text{mol B} = 9\,\text{mol B} \end{equation}\]
03

Calculating Moles of B for 2 mol of A

Repeating the process for 2 moles of A, we find that: \[\begin{equation} \frac{2\,\text{mol A}}{2\,\text{mol A}} \times 3\,\text{mol B} = 3\,\text{mol B} \end{equation}\]
04

Calculating Moles of B for 7 mol of A

Similarly, for 7 moles of A, the calculation is: \[\begin{equation} \frac{7\,\text{mol A}}{2\,\text{mol A}} \times 3\,\text{mol B} = 10.5\,\text{mol B} \end{equation}\]
05

Calculating Moles of B for 11 mol of A

Finally, for 11 moles of A: \[\begin{equation} \frac{11\,\text{mol A}}{2\,\text{mol A}} \times 3\,\text{mol B} = 16.5\,\text{mol B} \end{equation}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Chemical Reactions
Chemical reactions are processes where substances, known as reactants, transform into different substances called products. These transformations occur through the breaking and forming of chemical bonds. This is often visualized in a chemical equation, which represents the reactants on the left side and the products on the right.

For instance, in the reaction given by the equation \[2 \mathrm{A}+3 \mathrm{B} \longrightarrow 3 \mathrm{C}\], A and B are the reactants that combine to form the product C. The numbers in front of the chemical symbols are called stoichiometric coefficients. They indicate the proportions in which reactants and products participate in the reaction, reflecting the conservation of mass principle.
The Mole Concept in Chemistry
In chemistry, the term 'mole' is used to measure the quantity of a substance. One mole is equal to Avogadro's number (\(6.022 \times 10^{23}\)) of atoms, molecules, or other particles. This concept allows chemists to count particles by weighing them, as one mole of any element or compound has a mass that corresponds to its relative molecular (or atomic) mass in grams.

To tie this into our sample reaction, when we say '6 moles of A', we are referring to \(6 \times 6.022 \times 10^{23}\) particles of substance A. The mole concept is fundamental when it comes to quantifying substances in a chemical reaction and is crucial for stoichiometric calculations.
Understanding Molar Ratios
Molar ratios are derived from the coefficients of a balanced chemical equation and they reflect the proportions of the substances involved in the reaction. They are the heart of stoichiometry, as they enable us to convert between moles of different reactants and products.

In our exercise, the molar ratio between A and B is 2:3. This means that, ideally, two moles of A will react with three moles of B. Molar ratios allow us to predict the amount of reactants needed to completely react with a given quantity of another reactant, which is essential for practical applications such as chemical manufacturing.
Stoichiometric Calculations in Practice
Stoichiometric calculations use the mole concept and molar ratios to determine the quantitative aspects of chemical reactions. These calculations require a balanced chemical equation as a starting point. The process generally involves converting grams to moles, using the molar ratio to find the number of moles of another substance, and then, if necessary, converting back to grams.

For example, if we start with 6 moles of A, we can find out how many moles of B are required to completely react with A by setting up a proportion using the molar ratio. Our exercise demonstrates this step by step, where knowing that 2 moles of A react with 3 moles of B, we calculate the amount of B needed for different quantities of A. Understanding stoichiometric calculations allows students to predict the outcome of a reaction, a foundational skill in the field of chemistry.

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Most popular questions from this chapter

Consider the combustion of propane: $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Balance the reaction. (b) Divide all coefficients by the coefficient on propane, so that you have the reaction for the combustion of 1 mol of propane. (c) \(\Delta H_{\mathrm{rxn}}\) for the combustion of one mole of propane is \(-2219 \mathrm{~kJ}\). What mass of propane would you need to burn to generate \(5.0 \mathrm{MJ}\) of heat? (d) If propane costs about \(\$ 0.67 / \mathrm{L}\) and has a density of \(2.01 \mathrm{~g} / \mathrm{cm}^{3}\), how much would it cost to generate \(5.0 \mathrm{MJ}\) of heat by burning propane?

Consider the balanced equation: $$ \mathrm{SiO}_{2}(s)+3 \mathrm{C}(s) \longrightarrow \mathrm{SiC}(s)+2 \mathrm{CO}(g) $$ Complete the table with the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product formed. If the number of moles of a product is provided, fill in the required amount of each reactant to make that amount of product, as well as the amount of the other product that is made.

For the reaction shown, calculate how many moles of \(\mathrm{NH}_{3}\) form when each amount of reactant completely reacts. $$ 3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g) $$ (a) \(5.3 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (b) \(2.28 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (c) \(5.8 \times 10^{-2} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\) (d) \(9.76 \times 10^{7} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{H}_{4}\)

Consider the reaction between calcium oxide and carbon dioxide: $$ \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{CaCO}_{3}(s) $$ A chemist allows \(14.4 \mathrm{~g}\) of \(\mathrm{CaO}\) and \(13.8 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) to react. When the reaction is finished, the chemist collects \(19.4 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\). Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume there is more than enough of the other reactant. $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) \(4.6 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(4.6 \mathrm{~mol} \mathrm{O}_{2}\) (c) \(0.0558 \mathrm{~mol} \mathrm{C}_{3} \mathrm{H}_{8}\) (d) \(0.0558 \mathrm{~mol} \mathrm{O}_{2}\)

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