Question

Lead nitrate and potassium iodide react in aqueous solution to form a yellow precipitate of lead iodide. In one series of experiments, the masses of the two reactants were varied, but the total mass of the two was held constant at \(5.000 \mathrm{g}\). The lead iodide formed was filtered from solution, washed, dried, and weighed. The table gives data for a series of reactions. $$\begin{array}{lll} \hline & \text { Mass of Lead } & \text { Mass of Lead } \\ \text { Experiment } & \text { Nitrate, } g & \text { lodide, } g \\ \hline 1 & 0.500 & 0.692 \\ 2 & 1.000 & 1.388 \\ 3 & 1.500 & 2.093 \\ 4 & 3.000 & 2.778 \\ 5 & 4.000 & 1.391 \\ \hline \end{array}$$ (a) Plot the data in a graph of mass of lead iodide versus mass of lead nitrate, and draw the appropriate curve(s) connecting the data points. What is the maximum mass of precipitate that can be obtained? (b) Explain why the maximum mass of precipitate is obtained when the reactants are in their stoichiometric proportions. What are these stoichiometric proportions expressed as a mass ratio, and as a mole ratio? (c) Show how the stoichiometric proportions determined in part (b) are related to the balanced equation for the reaction.

Short answer

The maximum mass of the precipitate is obtained when the reactants are in their stoichiometric proportions because that's when all the reactants can fully react to produce the maximum possible amount of product. The stoichiometric proportions, expressed as a mass ratio and a mole ratio, can be calculated from the data at this point. These ratios should match with the coefficients in the balanced chemical equation for the reaction.

Step-by-step solution

Plot the data

Use the given data to plot a graph of mass of lead iodide (on the y-axis) against mass of lead nitrate (on the x-axis). Draw the curve that connects the data points.

Determine the maximum mass of precipitate

Look at the graph and find the highest point on the curve - this represents the maximum mass of lead iodide that can be obtained.

Explain why maximum mass of precipitate is obtained at stoichiometric proportions

The maximum mass of precipitate is obtained when the reactants are in their stoichiometric proportions because this is the point at which all the reactants are fully used up in the reaction, leading to the optimal amount of product.

Calculate the mass ratio

Find the stoichiometric proportions expressed as a mass ratio by dividing the mass of lead nitrate at which the maximum mass of lead iodide is formed by the mass of lead iodide obtained.

Calculate the mole ratio

Convert the masses of both lead nitrate and lead iodide to moles by dividing each by their respective molar masses. The ratio of these two values is the stoichiometric proportions expressed as a mole ratio.

Relate the stoichiometric proportions to the balanced equation

In the balanced equation for the reaction, the coefficients of the reactants and products represent the mole ratio in which they react/are formed. Verify that this ratio matches the mole ratio calculated in step 5 to demonstrate how the stoichiometric proportions are related to the balanced chemical equation.

Key concepts

Precipitation Reaction

In chemical reactions, a precipitation reaction is one where two soluble substances react to form an insoluble solid, known as a precipitate. This type of reaction occurs when two aqueous solutions containing soluble salts are mixed, resulting in the formation of an insoluble salt.
The reaction between lead nitrate and potassium iodide is a classic example. Here, lead nitrate ce{(Pb(NO3)2} and potassium iodide ce{(KI)} in solution react to form lead iodide ce{(PbI2)}, a yellow precipitate, and soluble potassium nitrate ce{(KNO3)}.
Precipitation reactions are crucial in laboratories for separating substances and for removing impurities from solutions. Pay close attention to the solubility rules to predict whether a precipitate will form during a reaction.

Stoichiometry

Stoichiometry is the study of the quantitative aspects of chemical reactions. It involves understanding the exact proportions of reactants and products involved in a chemical reaction.
In any chemical reaction, it is vital to know how much of each reactant is required to produce a desired amount of product without leaving an excess. This is where stoichiometry comes in. It allows chemists to predict the outcomes of reactions based on the balanced chemical equation, which presents the ratios of reactants and products.
By employing stoichiometry, one can calculate how much lead iodide can be formed when varying amounts of lead nitrate and potassium iodide are reacted, provided the reactions proceed to completion.

Mole Ratio

The mole ratio is derived from the balanced chemical equation of a reaction, indicating the proportion in which reactants combine and products form. It is a fundamental concept in stoichiometry.
For the reaction between lead nitrate and potassium iodide, the balanced equation is: \[ \text{Pb(NO}_3\text{)}_2 (aq) + 2 \text{KI} (aq) \rightarrow \text{PbI}_2 (s) + 2 \text{KNO}_3 (aq) \]From this equation, you can see that lead nitrate and potassium iodide react in a 1:2 mole ratio. This means one mole of lead nitrate reacts with two moles of potassium iodide to form one mole of lead iodide.
Understanding this ratio is crucial for calculating how much of each reactant you need to produce a specific amount of product.

Mass Ratio

The mass ratio refers to the proportion of the masses of each reactant needed to form a given mass of product. This ratio is useful in practical applications to ensure the most efficient use of materials.
To find the mass ratio for a reaction, you can begin by calculating the mole ratio and then converting moles to grams using the molar masses of the reactants and products.
Taking the reaction forming lead iodide as an example, you need to know the molar masses of lead nitrate and potassium iodide. Once you have those, you can find the mass of lead iodide generated for each gram of lead nitrate used, indicating the stoichiometric mass ratio.
This ratio helps determine the most efficient mix of reactants for maximum product yield.