What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO \(_{3}\) ) in 43.88 mL of a \(0.3842 M\) solution of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ?\) \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+4 \mathrm{KI} \longrightarrow 2 \mathrm{CuI}+\mathrm{I}_{2}+4 \mathrm{KNO}_{3}\)

When dealing with solutions in chemistry, molarity is a crucial concept to understand. It refers to the concentration of a solute in a solution. The formula to calculate molarity is quite straightforward: \[\begin{equation}M = \frac{\text{moles of solute}}{\text{liters of solution}}\end{equation}\]For instance, if a question asks you about the volume of a specific molarity (M) solution needed to obtain a certain amount of substance, you will first need to determine the number of moles of solute based on the molarity and then use the formula to find the volume.

Applying Molarity in Calculations

When you are given the volume and molarity of one solution and need to find out how much of another substance it reacts with, you apply the concept of molarity in two steps: First, finding the amount of substance in moles and then using stoichiometry to find how it relates to another reactant or product.For the homework problem provided, to find the volume of a 0.2089 M KI solution necessary to react with the Cu(NO3)2, we follow these steps:

• Calculate moles of Cu(NO3)2 using its volume and molarity.
• Use the reaction stoichiometry to find the corresponding moles of KI.
• Divide the moles of KI by its molarity to find the volume needed.

Following this method, you'll not only answer the exercise question accurately but also lay a strong foundation for understanding more complex molarity calculations. Remember, clear and simple steps with correct units are the secret to mastering molarity calculations.

In chemistry, reactions are described using chemical equations, which must be balanced to reflect the conservation of mass. Balancing a chemical equation means ensuring that the same number of atoms for each element appears on both sides of the equation. Learning to balance equations is vital before you can tackle stoichiometry problems.Often, simple patterns emerge that can guide the balancing process:

• Balance compounds containing multiple atoms of a single element last.
• Balance more complex molecules first, and leave single elements for last.
• If an element appears in more than one compound on the same side, it might be easier to balance last.

Steps for Balancing Equations

The chemical equation in our exercise is already balanced for us: \[\begin{equation}2 \text{Cu(NO}_{3})_{2} + 4 \text{KI} \rightarrow 2 \text{CuI} + \text{I}_{2} + 4 \text{KNO}_{3}\end{equation}\]It illustrates the reaction between copper(II) nitrate and potassium iodide. Here, stoichiometry is evident, as 2 moles of Cu(NO3)2 react with 4 moles of KI. Always double-check the balance of an equation before using it for stoichiometric calculations, as an unbalanced equation will lead to incorrect conclusions.

Reaction stoichiometry is the aspect of chemistry that deals with the quantitative relationships, or ratios, between reactants and products in a chemical reaction. Once a chemical equation is balanced, these ratios can be used to determine how much of a reactant is needed to create a certain amount of product or how much product can be formed from a given amount of reactant.

Applying Stoichiometry to Problems

In the provided exercise, once we know the moles of Cu(NO3)2, we use the stoichiometry of the reaction to determine how many moles of KI are needed. According to the balanced chemical equation, the ratio of KI to Cu(NO3)2 is 4:2, or simply 2:1.To solve a stoichiometry problem effectively:

• Start by writing the balanced chemical equation.
• Determine the molar ratio between the reactants and products you are interested in.
• Use the molar ratio to calculate the moles of one substance if the moles of another are known.

If you take these steps, you will approach stoichiometry problems with ease. The key is always to ensure you are working with a balanced equation and you understand the molar relationships it describes. With practice, reaction stoichiometry will change from a complex concept into a powerful tool that you can use with confidence.