Question

For each of the following pairs of combinations, indicate which one will produce the greater mass of solid product: a) \(105.5 \mathrm{~mL} 1.508 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(250.0 \mathrm{~mL}\) \(1.2075 \mathrm{M} \mathrm{KCl}\) or \(138.5 \mathrm{~mL} 1.469 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL} 2.115 \mathrm{M} \mathrm{KCl}\) b) \(32.25 \mathrm{~mL} 0.9475 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(92.75 \mathrm{~mL} 0.7750 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(52.50 \mathrm{~mL} 0.6810 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(39.50 \mathrm{~mL} 1.555 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) c) \(29.75 \mathrm{~mL} 1.575 \mathrm{M} \mathrm{AgNO}_{3}\) and \(25.00 \mathrm{~mL} 2.010 \mathrm{M}\) \(\mathrm{BaCl}_{2}\) or \(52.80 \mathrm{~mL} 2.010 \mathrm{M} \mathrm{AgNO}_{3}\) and \(73.50 \mathrm{~mL} 0.7500 \mathrm{M}\) \(\mathrm{BaCl}_{2}\)

Short answer

a) Scenario A; b) Scenario B; c) Scenario B.

Step-by-step solution

Determine Moles in Scenario A

Calculate the moles of each reactant for the reaction using the formula \(\text{moles} = \text{molarity} \times \text{volume in L}\). First, for \(105.5 \text{ mL of } \mathrm{Pb}(\mathrm{NO}_3)_2\) with molarity 1.508 M, we have:\[ \text{moles of } \mathrm{Pb}(\mathrm{NO}_3)_2 = 105.5 \times 10^{-3} \times 1.508 = 0.159 \text{ mol} \]For \(250.0 \text{ mL of } \mathrm{KCl}\) with molarity 1.2075 M, we have:\[ \text{moles of } \mathrm{KCl} = 250.0 \times 10^{-3} \times 1.2075 = 0.302 \text{ mol} \]

Identify Limiting Reactant in Scenario A

The reaction between \(\mathrm{Pb}(\mathrm{NO}_3)_2\) and \(\mathrm{KCl}\) produces \(\mathrm{PbCl}_2\) as the solid. The balanced equation is:\[ \mathrm{Pb}(\mathrm{NO}_3)_2 + 2\mathrm{KCl} \rightarrow \mathrm{PbCl}_2 + 2\mathrm{KNO}_3 \]From stoichiometry, 1 mole of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) requires 2 moles of \(\mathrm{KCl}\). Based on available moles, the need is:For 0.159 mol of \(\mathrm{Pb}(\mathrm{NO}_3)_2\), 0.318 mol of \(\mathrm{KCl}\) is required.Since only 0.302 mol \(\mathrm{KCl}\) is available, \(\mathrm{KCl}\) is the limiting reactant.

Determine Maximum Moles of PbCl2 in Scenario A

From the limiting reactant \(\mathrm{KCl}\), the moles of \(\mathrm{PbCl}_2\) formed is half the moles of \(\mathrm{KCl}\) because the molar ratio is 1:2. Thus:\[ \text{Moles of } \mathrm{PbCl}_2 = \frac{0.302}{2} = 0.151 \text{ mol} \]

Repeat Steps 1-3 for Scenario B

Calculate the moles for \(138.5 \text{ mL} \mathrm{Pb}(\mathrm{NO}_3)_2\) at 1.469 M and \(100.0 \text{ mL} \mathrm{KCl}\) at 2.115 M:\[ \text{moles of } \mathrm{Pb}(\mathrm{NO}_3)_2 = 138.5 \times 10^{-3} \times 1.469 = 0.203 \text{ mol} \]\[ \text{moles of } \mathrm{KCl} = 100.0 \times 10^{-3} \times 2.115 = 0.2115 \text{ mol} \]For the reaction, 0.203 mol of \(\mathrm{Pb}(\mathrm{NO}_3)_2\) requires 0.406 mol of \(\mathrm{KCl}\), thus \(\mathrm{KCl}\) is limiting as before, and the moles of \(\mathrm{PbCl}_2\) are:\[ \text{Moles of } \mathrm{PbCl}_2 = \frac{0.2115}{2} = 0.10575 \text{ mol} \]

Compare Moles of PbCl2

For part (a), compare the calculated moles of \(\mathrm{PbCl}_2\):- Scenario A: 0.151 mol- Scenario B: 0.10575 molScenario A produces more \(\mathrm{PbCl}_2\) solid.

Repeat Steps 1-5 for Part B

To solve part (b), follow the same method for mixtures of \(\mathrm{Na}_3\mathrm{PO}_4\) and \(\mathrm{Ca}(\mathrm{NO}_3)_2\) which yield \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\).For Scenario A:Moles of \(\mathrm{Na}_3\mathrm{PO}_4 = 32.25 \times 10^{-3} \times 0.9475 = 0.0306 \text{ mol} \)Moles of \(\mathrm{Ca}(\mathrm{NO}_3)_2 = 92.75 \times 10^{-3} \times 0.775 = 0.0719 \text{ mol} \)Balanced equation:\[ 3\mathrm{Ca}(\mathrm{NO}_3)_2 + 2\mathrm{Na}_3\mathrm{PO}_4 \rightarrow \mathrm{Ca}_3(\mathrm{PO}_4)_2 + 6\mathrm{NaNO}_3 \]0.0719 mol \(\mathrm{Ca}(\mathrm{NO}_3)_2\) needs 0.0479 mol \(\mathrm{Na}_3\mathrm{PO}_4\), \(\mathrm{Na}_3\mathrm{PO}_4\) is limiting, thus moles \(\mathrm{Ca}_3(\mathrm{PO}_4)_2 = \frac{0.0306}{2} = 0.0153\).For Scenario B:Moles of \(\mathrm{Na}_3\mathrm{PO}_4 = 52.50 \times 10^{-3} \times 0.681 = 0.03575 \text{ mol} \)Moles of \(\mathrm{Ca}(\mathrm{NO}_3)_2 = 39.50 \times 10^{-3} \times 1.555 = 0.0614 \text{ mol} \)For \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\):0.0614 mol \(\mathrm{Ca}(\mathrm{NO}_3)_2\) needs 0.0409 mol \(\mathrm{Na}_3\mathrm{PO}_4\), \(\mathrm{Na}_3\mathrm{PO}_4\) is limiting, thus moles \(\mathrm{Ca}_3(\mathrm{PO}_4)_2 = \frac{0.03575}{2} = 0.017875\).Compare: Scenario B makes more \(\mathrm{Ca}_3(\mathrm{PO}_4)_2\).

Part C Execution

Use the same method to solve for \(\mathrm{AgCl}\) produced from \(\mathrm{AgNO}_3\) and \(\mathrm{BaCl}_2\).For Scenario A:Moles of \(\mathrm{AgNO}_3 = 29.75 \times 10^{-3} \times 1.575 = 0.0469 \text{ mol} \)Moles of \(\mathrm{BaCl}_2 = 25.00 \times 10^{-3} \times 2.010 = 0.05025 \text{ mol} \)Balanced reaction:\[ 2\mathrm{AgNO}_3 + \mathrm{BaCl}_2 \rightarrow 2\mathrm{AgCl} + \mathrm{Ba(NO}_3)_2 \]0.05025 mol \(\mathrm{BaCl}_2\) needs 0.1005 mol \(\mathrm{AgNO}_3\), \(\mathrm{AgNO}_3\) is limiting, moles \(\mathrm{AgCl}=\frac{0.0469}{1}=0.0469\).For Scenario B:Moles of \(\mathrm{AgNO}_3 = 52.80 \times 10^{-3} \times 2.010 = 0.1061 \text{ mol} \)Moles of \(\mathrm{BaCl}_2 = 73.50 \times 10^{-3} \times 0.750 = 0.0551 \text{ mol} \)0.0551 mol \(\mathrm{BaCl}_2\) needs 0.1102 mol \(\mathrm{AgNO}_3\), \(\mathrm{AgNO}_3\) is limiting, moles \(\mathrm{AgCl}=\frac{0.1061}{1}=0.1061\).Compare: Scenario B makes more \(\mathrm{AgCl}\).

Key concepts

Stoichiometry

Stoichiometry is a fundamental aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. This allows chemists to determine the amount of products that will be formed in a reaction based on the amount of reactants used. The key to understanding stoichiometry is the balanced chemical equation, which provides the ratio of reactants to products. For example, in the equation \( \mathrm{Pb(NO}_3)_2 + 2\mathrm{KCl} \rightarrow \mathrm{PbCl}_2 + 2\mathrm{KNO}_3 \), it indicates that one mole of lead nitrate reacts with two moles of potassium chloride to produce one mole of lead chloride and two moles of potassium nitrate.
To solve stoichiometric problems, follow these steps:

• Convert all reactant quantities (often given in grams or milliliters) into moles using their molar masses or given molarity and volume.
• Use the balanced chemical equation to find the ratio and calculate how many moles of each product one can obtain based on the reactant quantities.
• Identify the limiting reactant, which determines the maximum amount of product that can be formed.
• Calculate the theoretical yield using stoichiometry, predicting the amount of product.

Stoichiometry is essential for predicting yields in reactions, both in laboratory and industrial settings, thus saving resources, time, and energy.

Molarity

Molarity is a way of expressing the concentration of a solution. It is defined as the number of moles of solute (the substance being dissolved) per liter of solution. The unit is usually represented as M, where 1 M = 1 mole per liter. Understanding molarity is crucial because it directly relates to stoichiometry and helps determine how much of a substance is present in a given volume of solution.
For instance, in the original exercise, molarity was used to calculate the number of moles of each reactant. The formula used is: \[ \text{Moles} = \text{Molarity} \times \text{Volume in Liters} \]This formula allows conversion from volume and concentration (molarity) to moles, which can then be used in stoichiometric calculations. For example, calculating moles for 105.5 mL of \( \mathrm{Pb(NO}_3)_2 \) with a molarity of 1.508 M gives you 0.159 moles.
Using molarity helps ensure that reactions are carried out with the correct proportions of substances, which is critical for achieving the desired chemical transformation efficiently and safely.

Balanced Chemical Equation

A balanced chemical equation is an expression that represents the relationship between the reactants and products in a chemical reaction. Each chemical equation must be balanced because of the law of conservation of mass, which states that matter cannot be created nor destroyed. Therefore, the number of atoms for each element must be the same on both sides of the equation.
To balance a chemical equation:

• Identify how many atoms of each element are present on both sides of the equation.
• Use coefficients to adjust the number of molecules so that the number of each type of atom is equal on both sides.
• Check to ensure all coefficients are the smallest possible integers and that all elements are balanced.

In the reaction between \( \mathrm{Pb(NO}_3)_2 \) and \( \mathrm{KCl} \) to form \( \mathrm{PbCl}_2 \), the balanced equation is \( \mathrm{Pb(NO}_3)_2 + 2\mathrm{KCl} \rightarrow \mathrm{PbCl}_2 + 2\mathrm{KNO}_3 \). This indicates that one mole of lead nitrate reacts with two moles of potassium chloride to produce one mole of lead chloride and two moles of potassium nitrate. This balance ensures that calculations for stoichiometry are accurate.

Precipitation Reaction

A precipitation reaction occurs when two solutions, containing soluble salts, are mixed together, resulting in the formation of an insoluble salt (precipitate). This kind of reaction is characterized by the formation of a solid product, which settles out of the solution. Precipitation reactions are common in qualitative analysis to identify the presence of certain ions within a solution.
The reaction between \( \mathrm{Pb(NO}_3)_2 \) and \( \mathrm{KCl} \) is a classic example of a precipitation reaction. When these two are mixed, they form \( \mathrm{PbCl}_2 \), a solid precipitate, along with \( \mathrm{KNO}_3 \), which remains in solution. The balanced chemical equation is:\[ \mathrm{Pb(NO}_3)_2 + 2\mathrm{KCl} \rightarrow \mathrm{PbCl}_2(s) + 2\mathrm{KNO}_3(aq) \]
This equation shows that the lead chloride is the precipitate, denoted by (s) for solid, while the potassium nitrate remains aqueous (aq). Such reactions are quite useful in various industrial and laboratory processes, including water purification and the synthesis of chemicals. Recognizing precipitation reactions is essential in predicting the outcomes of chemical mixtures.