Question

For each of the following unbalanced reactions, suppose exactly 5.00 moles of each reactant are taken. Determine which reactant is limiting, and also determine what mass of the excess reagent will remain after the limiting reactant is consumed. For cach reaction, solve the problem three ways: i. Set up and use Before-Change-After (BCA) tables. ii. Compare the moles of reactants to see which runs out first. iii. Consider the amounts of products that can be formed by completcly consuming cach reactant. a. \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) b. \(\operatorname{AgNO}_{3}(a q)+\mathbf{A l}(s) \rightarrow \mathbf{A}_{\mathbf{g}}(s)+\mathbf{A l}\left(\mathrm{NO}_{3}\right)_{3}(a q)\)

Short answer

For the first reaction, we determined that H₂O is the limiting reactant, and the mass of excess CaC₂ remaining after the reaction is 160.25 g. For the second reaction, AgNO₃ is the limiting reactant, and the mass of excess Al remaining after the reaction is 89.82 g.

Step-by-step solution

Calculate Moles of Reactants and Products

Calculate and record the moles of each reactant before, during, and after the reaction: - Before: 5.00 moles CaC₂ and 5.00 moles H₂O - Change: -x moles CaC₂ and -2x moles H₂O - After: (5.00 - x) moles CaC₂ and (5.00 - 2x) moles H₂O

Identify the Limiting Reactant

From the above data, we can set up a BCA table and identify the limiting reactant. Since 1 mole of CaC₂ reacts with 2 moles of H₂O, we can see that there are not enough moles of H₂O to react with all the CaC₂. Hence, H₂O is the limiting reactant.

Calculate Excess Reactant Remaining

Now, we will solve for x using the limiting reactant: \(5 - 2x = 0 \Rightarrow x = 2.5\). Therefore, 2.5 moles of CaC₂ remain after the reaction, and we can convert this to mass using the molar mass: \(\mathrm{mass}=\mathrm{moles} \times \mathrm{molar\, mass} = 2.5\, \mathrm{moles} \times 64.10\, \mathrm{g/mol} = 160.25\, \mathrm{g}\). ii. Compare the moles of reactants

Calculate Stoichiometric Ratio

Calculate the stoichiometric ratio of moles of H₂O to moles of CaC₂: \(\frac{5\, \mathrm{moles\, H_{2}O}}{5\, \mathrm{moles\, CaC_{2}}} = 1\) Since 1 mole of CaC₂ reacts with 2 moles of H₂O, it is clear from the stoichiometric ratio that we don't have enough moles of H₂O, making it the limiting reactant. It also tells us that we will have an excess of CaC₂ remaining after the reaction. We already calculated the remaining mass of CaC₂ in method i. iii. Amounts of products formed

Amount of Products Formed

Let's consider the amount of product formed by completely consuming each reactant: - If 5 moles of CaC₂ are completely consumed, we will get 5 moles of C₂H₂ according to the balanced equation. - If 5 moles of H₂O are completely consumed, we will get 2.5 moles of C₂H₂. Since lesser moles of C₂H₂ are formed when H₂O is completely consumed, it means H₂O is the limiting reactant. Excess reactant remaining can be calculated as in method i (resulting in 160.25 g of CaC₂ remaining). ##b. Reaction 2: \(\mathrm{AgNO}_{3}(aq)+\mathrm{Al}(s) \rightarrow \mathrm{Ag}(s)+\mathrm{Al}(\mathrm{NO}_{3})_{3}(aq)\)## First, we need to balance the reaction: \(3\mathrm{AgNO}_{3}(aq)+1\mathrm{Al}(s) \rightarrow 3\mathrm{Ag}(s)+1\mathrm{Al}(\mathrm{NO}_{3})_{3}(aq)\) Now, let's solve the problem using three different methods. i. BCA tables

Calculate Moles of Reactants and Products

Calculate and record the moles of each reactant before, during, and after the reaction: - Before: 5.00 moles AgNO₃ and 5.00 moles Al - Change: -3x moles AgNO₃ and -x moles Al - After: (5.00 - 3x) moles AgNO₃ and (5.00 - x) moles Al

Identify the Limiting Reactant

From the above data, we can see that there are not enough moles of AgNO₃ to react with all the Al because 3 moles of AgNO₃ are required to react with 1 mole of Al. Hence, AgNO₃ is the limiting reactant.

Calculate Excess Reactant Remaining

Now, we will solve for x using the limiting reactant: \(5 - 3x = 0 \Rightarrow x = 1.67\). Therefore, 3.33 moles of Al remain after the reaction, and we can convert this to mass using the molar mass: \(\mathrm{mass}=\mathrm{moles} \times \mathrm{molar\, mass} = 3.33\, \mathrm{moles} \times 26.98\, \mathrm{g/mol} = 89.82\, \mathrm{g}\). ii. Compare the moles of reactants

Calculate Stoichiometric Ratio

Calculate the stoichiometric ratio of moles of AgNO₃ to moles of Al: \(\frac{5\, \mathrm{moles\, AgNO_{3}}}{5\, \mathrm{moles\, Al}} = 1\) Since 1 mole of Al reacts with 3 moles of AgNO₃, it is clear from the stoichiometric ratio that we don't have enough moles of AgNO₃, making it the limiting reactant. It also tells us that we will have an excess of Al remaining after the reaction. We already calculated the remaining mass of Al in method i. iii. Amounts of products formed

Amount of Products Formed

Let's consider the amount of product formed by completely consuming each reactant: - If 5 moles of AgNO₃ are completely consumed, we will get 1.67 moles of Ag according to the balanced equation. - If 5 moles of Al are completely consumed, we will get 15 moles of Ag. Since lesser moles of Ag are formed when AgNO₃ is completely consumed, it means AgNO₃ is the limiting reactant. Excess reactant remaining can be calculated as in method i (resulting in 89.82 g of Al remaining).

Key concepts

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculations that use molar ratios from balanced chemical equations to determine the amount of reactants needed or products formed.

Imagine you're baking a cake and the recipe calls for two eggs and one cup of flour. If you start with five eggs and five cups of flour, knowing how to 'scale' the recipe helps you figure out how much cake you can make without running out of ingredients. Stoichiometry does something similar; it tells us how much product we can expect from a chemical reaction given a certain amount of each reactant.

For instance, when comparing moles of reactants to understand which one will run out first, you're practicing stoichiometry. If we look at the exercise's Reaction a, we see the following reaction:

1. Before: 5.00 moles of each reactant.
2. Compare quantities: Since 1 mole of CaC₂ needs 2 moles of H₂O to react fully, we can immediately see that H₂O will be the limiting reactant.
3. After: We're left with an excess of CaC₂ because there wasn't enough H₂O to react with all the CaC₂.

Stoichiometry helps us predict this outcome and assess the limits of a reaction.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking of old and formation of new chemical bonds. Reactants convert to products, and understanding this process is fundamental in chemistry. When balancing chemical equations, we ensure that the number of atoms for each element is conserved in the reaction. It's like making sure that all the pieces of a puzzle are used without any leftovers.

In the context of our textbook solution, Reaction b is an example of a single displacement reaction: Silver nitrate (AgNO₃) reacts with aluminum (Al) to form silver (Ag) and aluminum nitrate (Al(NO₃)₃). We must first balance the chemical equation to ensure the law of conservation of mass is obeyed:

3 AgNO₃ + Al → 3 Ag + Al(NO₃)₃

This balanced equation then guides us to determine the limiting reactant and the quantities of products that can be formed. When analyzing this reaction, we can apply stoichiometric principles to determine that we would need 3 moles of AgNO₃ for every mole of Al to ensure neither is in excess. However, since we only have equal moles of each, AgNO₃ will be the limiting reactant.

Mole Concept

The mole concept is a way to quantify substances in chemistry based on the fact that one mole contains exactly 6.022 x 10²³ particles, often referred to as Avogadro's number. This concept allows chemists to measure substances using a common unit, making calculations and laboratory work feasible.

It’s like counting eggs in dozens; just as a dozen represents a count of 12, a mole represents a much larger count of particles. In a chemical reaction, it’s essential to know the number of moles of each reactant you have to begin with, which enables you to predict how much product you can produce.

Referring back to the exercise, by identifying the limiting reactant, we determine which reactant will be entirely consumed and therefore restrict the amount of product produced. By thinking in terms of moles, we can easily compare the proportion of reactants, as shown in both reactions a and b. Remember the cake analogy? Here, moles help us understand if we have a 'dozen' or just a 'half dozen' of our ingredients. The remaining mass of the excess reactant, like the leftover flour, is simply the amount you started with minus what was used to react with the limiting reactant.