Question

When \(200.0 \mathrm{~mL}\) of \(0.10 M \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KOH}, \mathrm{a}\) precipitate forms. a. How many moles of precipitate can form in this reaction? b. Calculate the concentration of \(\mathrm{Zn}^{2+}\) ions in the final solution after precipitate formation is complete.

Short answer

a. The moles of precipitate that can form in this reaction is \(0.020 \: moles\) of Zn(OH)2. b. The concentration of Zn^2+ ions in the final solution after precipitate formation is complete is 0 M.

Step-by-step solution

Write the balanced chemical equation

The given reaction is: \[Zn(NO_3)_2 + 2KOH \rightarrow Zn(OH)_2↓ + 2KNO_3\] In this reaction, Zn(OH)_2 is the precipitate that forms.

Calculate the moles of reactants

Now we'll use the molarity and volume of each reactant to find the moles for each. For Zn(NO3)2, \(Moles = Molarity \times Volume\): Moles = (0.10 M) × (200.0 mL) × \(\frac{1L}{1000 ml}\) = 0.020 moles Zn(NO3)2 For KOH: Moles = (0.10 M) × (100.0 mL) × \(\frac{1L}{1000 ml}\) = 0.010 moles KOH

Determine the limiting reactant

We'll use the stoichiometry from the balanced chemical equation and the moles of the reactants to determine the limiting reactant. The stoichiometric ratio is 1:2:1:2 for Zn(NO3)2:KOH:Zn(OH)2:KNO3. We have 0.020 moles of Zn(NO3)2 and 0.010 moles of KOH. Therefore, the ratio is: (0.020 moles Zn(NO3)2) / (1) = (0.010 moles KOH) / (2) The ratio is equal, so both reactants are limiting reactants.

Calculate the moles of precipitate (Zn(OH)2) formed

As both reactants are limiting reactants, we can use the moles of either reactant to find the moles of Zn(OH)2 produced: \[Moles(Zn(OH)_{2}) = Moles(Zn(NO_3)_2)\] Moles of Zn(OH)2 = 0.020 moles a. The moles of precipitate that can form in this reaction is \(0.020 \: moles\) of Zn(OH)2.

Calculate the concentration of Zn^2+ ions in the final solution

As all of the zinc ions will form the Zn(OH)2 precipitate, there are no Zn^2+ ions left in the final solution. b. The concentration of Zn^2+ ions in the final solution after precipitate formation is complete is 0 M.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is like a recipe in chemistry. It shows us exactly how reactants interact to form products, ensuring that the number of atoms of each element is the same on both sides of the equation.
For the given reaction, the balanced equation is:

• \[Zn(NO_3)_2 + 2KOH \rightarrow Zn(OH)_2↓ + 2KNO_3\]

It illustrates how one molecule of zinc nitrate reacts with two molecules of potassium hydroxide to form one molecule of zinc hydroxide precipitate and two molecules of potassium nitrate.
The key idea here is the stoichiometry, which is the ratio of reactants to products. This balanced equation tells us that for every 1 mole of \( Zn(NO_3)_2 \), we need 2 moles of \( KOH \) to react completely. Understanding this ratio is crucial for predicting the amounts of products formed in a reaction.

Limiting Reactant

The limiting reactant is the substance that is entirely consumed first in a chemical reaction, determining the maximum amount of product that can be formed. To identify the limiting reactant, we look at the moles available and the stoichiometry of the balanced equation.
In this exercise, we have:

• 0.020 moles of \( Zn(NO_3)_2 \)
• 0.010 moles of \( KOH \)

According to the balanced equation, it takes 2 moles of \( KOH \) to react with 1 mole of \( Zn(NO_3)_2 \). Here, we have exactly twice as few moles of \( KOH \) as required for \( Zn(NO_3)_2 \), so we see that both are effectively the limiting reactants in the context provided.
This means they will be used up completely at the same time and determine how much \( Zn(OH)_2 \) can be formed. Recognizing which reactant limits the reaction helps in calculating the exact amount of product possible.

Mole Calculations

Mole calculations allow us to convert between amounts of substances in a reaction. This is essential for determining how much product a certain amount of reactant can produce. We typically use molarity (concentration) and volume to find the moles of each reactant in a solution.
For instance, in calculating the moles of \( Zn(NO_3)_2 \):

• Use the formula: \( \text{Moles} = \text{Molarity} \times \text{Volume} \)
• \( 0.10 \, \text{M} \times 200.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.020 \, \text{moles} \)

Similarly for \( KOH \):

• \( 0.10 \, \text{M} \times 100.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.010 \, \text{moles} \)

Using these mole calculations, we can predict the formation of \( 0.020 \, \text{moles} \) of \( Zn(OH)_2 \), the precipitate. Mastering these calculations gives us the ability to translate laboratory conditions into a quantitative analysis of reaction components.