Question

What volume of 0.779 M \(\mathrm{NaCl}\) will react with \(40.8 \mathrm{~mol}\) of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) ? \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \rightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3}\)

Short answer

104.75 L

Step-by-step solution

Understand the Reaction Equation

The balanced chemical equation provided is: \[ \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \rightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \].This tells us that one mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) reacts with two moles of \(\mathrm{NaCl}\).

Determine Moles of NaCl Required

We know that 40.8 moles of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) are reacting. From the balanced equation, two moles of \(\mathrm{NaCl}\) are needed per mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\). Therefore, the moles of \(\mathrm{NaCl}\) required are:\[ 40.8 \text{ moles of } \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \times 2 \frac{\text{mole } \mathrm{NaCl}}{\text{mole } \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}} = 81.6 \text{ moles of } \mathrm{NaCl} \]

Calculate Volume of NaCl Solution

The molarity \(M\) of the \(\mathrm{NaCl}\) solution is 0.779 M, which means 0.779 moles per liter. To find the volume \(V\) needed, use the formula: \[ M = \frac{n}{V} \Rightarrow V = \frac{n}{M} \]Substituting the moles and molarity:\[ V = \frac{81.6 \text{ moles }}{0.779 \text{ moles/L}} \approx 104.75 \text{ L} \]

Report the Volume Solution

The volume of \(0.779 \text{ M } \mathrm{NaCl}\) solution required to react with 40.8 moles of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) is approximately 104.75 liters.

Key concepts

Molarity

Molarity is a fundamental concept in chemistry, often used to describe the concentration of a solution. It is defined as the number of moles of a solute per liter of solution. Understanding molarity is essential for calculating how much of a particular substance is needed in a reaction.
When you see a molarity value like 0.779 M for NaCl, it means there are 0.779 moles of sodium chloride dissolved in one liter of solution. Knowing this, you can figure out how much of a solution you would need for a reaction. For instance, if you need a certain amount of moles, you can rearrange the molarity formula to solve for volume:

• Use the formula: \[ M = \frac{n}{V} \] Where \( M \) is molarity, \( n \) is the number of moles, and \( V \) is the volume in liters.
• Rearrange the formula to find volume: \[ V = \frac{n}{M} \]

So when working on chemistry problems, knowing molarity gives you the power to precisely control the concentration of your solutions.

Balanced Chemical Equation

A balanced chemical equation is crucial in understanding chemical reactions, as it shows the relationship between reactants and products in a reaction. Balance means that there are equal numbers of each type of atom on both sides of the equation, which reflects the Law of Conservation of Mass.
In the equation for our reaction: \[ \mathrm{Pb} \left( \mathrm{NO}_{3} \right)_{2} + 2 \mathrm{NaCl} \rightarrow \mathrm{PbCl}_{2} + 2 \mathrm{NaNO}_{3} \] it is balanced because the number of atoms of each element is the same on both sides:

• 1 lead (Pb) atom
• 2 sodium (Na) atoms
• 2 nitrate (NO₃) groups
• 2 chloride (Cl) atoms

Balanced equations help us determine the stoichiometric ratios. They tell us how many moles of each substance are involved in the reaction. For example, the equation tells us that 1 mole of lead(II) nitrate reacts with 2 moles of sodium chloride. These ratios are key to calculating how much of each reactant is needed, as demonstrated in step-by-step solutions.

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products. They are the core events in chemistry, representing the interactions that lead to new substances. Understanding this process helps in predicting reaction outcomes and calculating needed reagents.
The particular reaction here is a double displacement reaction, where components of two compounds exchange places to form two new compounds: \[ \mathrm{Pb} \left( \mathrm{NO}_{3} \right)_{2} + 2 \mathrm{NaCl} \rightarrow \mathrm{PbCl}_{2} + 2 \mathrm{NaNO}_{3} \] This type of reaction often occurs in aqueous solutions and leads to the formation of a precipitate, a solid formed out of solution.

• Lead(II) chloride \((\mathrm{PbCl}_{2})\) is the precipitate in this reaction.
• The driving force for the reaction is the formation of this insoluble product, removing ions from the solution.

Recognizing the type of chemical reaction helps predict the results and compute quantities needed, just as balancing equations helps maintain consistency with the laws of chemical stoichiometry.