Consider the unbalanced equation for the reaction of solid lead with silver nitrate: $$ \mathrm{Pb}(s)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{Ag}(s) $$ (a) Balance the equation. (b) How many moles of silver nitrate are required to completely react with \(9.3 \mathrm{~mol}\) of lead? (c) How many moles of Ag are formed by the complete reaction of \(28.4\) mol of \(\mathrm{Pb}\) ?

Short Answer

Expert verified
The balanced equation is: Pb(s) + 2AgNO3(aq) -> Pb(NO3)2(aq) + 2Ag(s). For (b), 18.6 mol AgNO3 are required for 9.3 mol Pb. For (c), 56.8 mol Ag are formed from 28.4 mol Pb.

Step by step solution

01

Balancing the Lead Atoms

In the reaction, there is 1 lead atom on the reactant side and 1 lead atom on the product side. So, the lead atoms are already balanced.
02

Balancing the Silver Atoms

There is 1 silver atom (in AgNO3) on the reactant side and 1 silver atom (Ag) on the product side. Therefore, silver atoms are also balanced.
03

Balancing the Nitrate Ions

There is 1 nitrate ion in AgNO3 on the reactant side, but there are 2 nitrate ions in Pb(NO3)2 on the product side. To balance the nitrate ions, you need 2 moles of AgNO3 for each mole of Pb(NO3)2. This means you should put a coefficient of 2 in front of AgNO3 to balance the nitrate ions.
04

Final Balanced Equation

The balanced equation is: \[ \mathrm{Pb}(s) + 2\mathrm{AgNO}_{3}(aq) \longrightarrow \mathrm{Pb(NO}_{3})_{2}(aq) + 2\mathrm{Ag}(s) \]
05

Calculating Moles of AgNO3 for Reaction with Pb

Using the stoichiometry from the balanced equation in Step 4, 2 moles of AgNO3 are required to react with 1 mole of Pb. Therefore, to react with 9.3 moles of Pb: \[ 2 \cdot 9.3 \, \mathrm{mol} \, \mathrm{AgNO}_{3} = 18.6 \, \mathrm{mol} \, \mathrm{AgNO}_{3}\]
06

Calculating Moles of Ag Formed from Pb Reaction

From the balanced equation, 2 moles of Ag are produced for every mole of Pb that reacts. Using this stoichiometry with 28.4 mol of Pb: \[ 2 \cdot 28.4 \, \mathrm{mol} \, \mathrm{Ag} = 56.8 \, \mathrm{mol} \, \mathrm{Ag}\]

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is a foundational skill in chemistry that ensures the law of conservation of matter is adhered to. This law states that matter cannot be created or destroyed. When balancing equations, our goal is to make sure the number of atoms of each element is equal on both the reactant and product sides.

For instance, when balancing the equation \[ \text{Pb}(s) + \text{AgNO}_{3}(aq) \longrightarrow \text{Pb(NO}_{3})_{2}(aq) + \text{Ag}(s) \], we observe each element or compound to ensure that they match up. Starting with the most complex molecule, typically a good practice, we work through each element to make sure it balances. The process might involve placing coefficients — whole numbers placed in front of compounds — to ensure the number of atoms of each element in the reactants equals those in the products. In our case, we ended up with the balanced equation \[ \text{Pb}(s) + 2\text{AgNO}_{3}(aq) \longrightarrow \text{Pb(NO}_{3})_{2}(aq) + 2\text{Ag}(s) \].
Mole-to-Mole Ratios
The mole-to-mole ratio is derived from the coefficients of a balanced chemical equation. It tells us how many moles of one substance will react with or produce a certain number of moles of another substance. A clear understanding of mole-to-mole ratios is crucial for precisely predicting the amounts of reactants or products involved in a chemical reaction.

In the solved exercise, the balanced equation indicates a mole-to-mole ratio of 1 mole of Pb to 2 moles of AgNO3 (\(1:\text{Pb}=2:\text{AgNO}_{3}\)) for the reactants and 1 mole of Pb to 2 moles of Ag (\(1:\text{Pb}=2:\text{Ag}\)) for the products. Using this ratio, it is straightforward to predict that to react with 9.3 moles of \(\text{Pb}\), one would need 18.6 moles of \(\text{AgNO}_{3}\).
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction, using the mole-to-mole ratios derived from a balanced equation. It tells us how much product can be formed from given reactants or how much reactants are needed to produce a certain amount of product, based on those proportions.

By following reaction stoichiometry, we can, for example, determine that 28.4 moles of Pb will produce 56.8 moles of Ag, using the molecule-to-molecule ratio expected from the reaction.

Considering Conversion Factors

In a real-world scenario, such calculations often incorporate additional steps to convert between units, such as grams to moles, using the molar mass as a conversion factor. This enables precise measurements in the lab and industrial processes.
Solving Stoichiometry Problems
When solving stoichiometry problems, students should follow a structured approach to avoid mistakes and ensure accuracy:

1. Start by writing down the unbalanced chemical equation.
2. Balance the equation using coefficients to get equal numbers of atoms of each element on both sides.
3. Determine the mole ratios from the balanced equation.
4. Use conversion factors as needed, such as molar masses, to translate between grams and moles or liters, if dealing with gases at STP (Standard Temperature and Pressure).
5. Apply the mole ratios to calculate the moles (or mass or volume) of reactants or products as required by the problem.

Utilizing Dimensional Analysis

Dimensional analysis, a mathematical method that employs conversion factors to move from one unit to another, is an indispensable tool in solving stoichiometry problems. This method prevents missteps by ensuring that units cancel properly and calculations are consistent with the mole ratios.

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Most popular questions from this chapter

Consider the equation for the combustion of acetone \(\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)\), the main ingredient in nail polish remover: $$ \mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)+4 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) } \\ {\Delta H_{\text {nxn }}=-1790 \mathrm{~kJ}} \end{array} $$ If a bottle of nail polish remover contains \(155 \mathrm{~g}\) of acetone, how much heat is released by its complete combustion?

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Consider the generic reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C} \Delta H_{\mathrm{rxn}}=-125 \mathrm{~kJ} $$ Determine the amount of heat emitted when each amount of reactant completely reacts (assume that there is more than enough of the other reactant). (a) \(2 \mathrm{~mol} \mathrm{~A}\) (b) \(3 \mathrm{~mol} \mathrm{~A}\) (c) \(3 \mathrm{~mol} \mathrm{~B}\) (d) \(5 \mathrm{~mol} \mathrm{~B}\)

For the reaction shown, calculate the theoretical yield of the product in grams for each of the initial quantities of reactants. $$ \mathrm{Ti}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{TiF}_{4}(s) $$ (a) \(1.0 \mathrm{~g} \mathrm{Ti} ; 1.0 \mathrm{~g} \mathrm{~F}_{2}\) (b) \(4.8 \mathrm{~g} \mathrm{Ti} ; 3.2 \mathrm{~g} \mathrm{~F}_{2}\) (c) \(0.388 \mathrm{~g} \mathrm{Ti} ; 0.341 \mathrm{~g} \mathrm{~F}_{2}\)

Consider the generic reaction: $$ \mathrm{A}+2 \mathrm{~B} \longrightarrow \mathrm{C} \Delta H_{\mathrm{ran}}=-55 \mathrm{~kJ} $$ Determine the amount of heat emitted when each amount of reactant completely reacts (assume that there is more than enough of the other reactant). (a) \(1 \mathrm{~mol} \mathrm{~A}\) (b) 2 mol \(\mathrm{A}\) (c) \(1 \mathrm{~mol} \mathrm{~B}\) (d) \(2 \mathrm{~mol} \mathrm{~B}\)

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