Problem 69 - General Chemistry: Principles and Modern Applications

Question

Ammonia can be generated by heating together the solids \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2} . \mathrm{CaCl}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are also formed. (a) If a mixture containing \(33.0 \mathrm{g}\) each of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) is heated, how many grams of \(\mathrm{NH}_{3}\) will form? (b) Which reactant remains in excess, and in what mass?

Step-by-Step Solution

Answer

a) 10.5 g of \( NH_{3} \) will form. b) \( Ca(OH)_{2} \) remains in excess with a mass of 10.1 g.

Step 1: Balancing the Chemical Equation

The first step is to write down the chemical reaction and balance it: \[\mathrm{NH}_{4} \mathrm{Cl} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{NH}_{3} + \mathrm{CaCl}_{2} + \mathrm{H}_{2} \mathrm{O}\] The balanced equation is: \[2\mathrm{NH}_{4} \mathrm{Cl} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow 2\mathrm{NH}_{3} + \mathrm{CaCl}_{2} + 2\mathrm{H}_{2} \mathrm{O}\]

Step 2: Calculation of Moles

The second step is to calculate the number of moles of reactants using the molar masses. For \(NH_{4}Cl\), the molar mass is 53.49 g/mol, and for \(Ca(OH)_{2}\), the molar mass is 74.09 g/mol. So, the moles of \( NH_{4}Cl \( are 33.0 g / 53.49 g/mol = 0.617 mol. For \( Ca(OH)_{2} \), the calculated moles are: 33.0 g / 74.09 g/mol = 0.445 mol.

Step 3: Identification of Limiting Reactant

In the balanced chemical equation, the stoichiometric ratio between \( NH_{4}Cl \) and \( Ca(OH)_{2} \) is 2:1. This means that two moles of \( NH_{4}Cl \) react with each mole of \( Ca(OH)_{2} \). However, we have less than 2 times the amount of \( NH_{4}Cl \) than \( Ca(OH)_{2} \), so \( NH_{4}Cl \) will be entirely consumed first. Therefore, \( NH_{4}Cl \) is the limiting reactant.

Step 4: Mass of Ammonia Formed

Based on the stoichiometry of the reaction, two moles of ammonia are produced for every two moles of \( NH_{4}Cl \) used. Since we start with 0.617 moles of \( NH_{4}Cl \), this means we form 0.617 moles of \( NH_{3} \). The molar mass of \( NH_{3} \) is 17.03 g/mol, so the mass of \( NH_{3} \) produced is 0.617 mol * 17.03 g/mol = 10.5 g.

Step 5: Calculating Excess Reactant Mass

The stoichiometry also tells us that for every 2 moles of \( NH_{4}Cl \) that react, 1 mole of \( Ca(OH)_{2} \) is needed. Since we used all 0.617 moles of \( NH_{4}Cl \), this means we used 0.617 / 2 = 0.3085 mol of \( Ca(OH)_{2} \). If we started with 0.445 mol, this means we have 0.445 - 0.3085 = 0.1365 mol of \( Ca(OH)_{2} \) left. The molar mass of \( Ca(OH)_{2} \) is 74.09 g/mol, so the mass of \( Ca(OH)_{2} \) remaining is 0.1365 mol * 74.09 g/mol = 10.1 g.

Key Concepts

Limiting Reactant

A limiting reactant in a chemical reaction is the substance that runs out first, stopping the reaction from continuing and limiting the amount of product that can be produced. Simply put, it determines how far a reaction can go and how much product can be formed.

To determine the limiting reactant, you'll need to compare the mole ratio of the reactants used in the chemical process with the ratio of the balanced equation. In the provided chemical reaction:

• he chemical equation tells us that 2 moles of \( \mathrm{NH}_{4} \mathrm{Cl} \) react with 1 mole of \( \mathrm{Ca} (\mathrm{OH})_{2} \).
• In our example, we have fewer moles of \( \mathrm{NH}_{4} \mathrm{Cl} \) than this ratio requires.
  
  
• Thus, \( \mathrm{NH}_{4} \mathrm{Cl} \) is the limiting reactant. It will deplete first and determines the maximum amount of ammonia \( \mathrm{NH}_{3} \) that can be produced.

Understanding which reactant is limiting is crucial because it helps us predict the outcome of the reaction accurately.

Chemical Equation Balancing

Balancing a chemical equation is necessary to ensure that the same number of each type of atom is present on both sides of the equation. This follows the Law of Conservation of Mass, stating that mass can neither be created nor destroyed in a chemical reaction. So, when you look at the unbalanced equation:

• \( \mathrm{NH}_{4} \mathrm{Cl} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{NH}_{3} + \mathrm{CaCl}_{2} + \mathrm{H}_{2} \mathrm{O} \)

You want to balance it by ensuring both sides have the same type and number of atoms.

After balancing, the equation becomes:

• \( 2\mathrm{NH}_{4} \mathrm{Cl} + \mathrm{Ca}(\mathrm{OH})_{2} \rightarrow 2\mathrm{NH}_{3} + \mathrm{CaCl}_{2} + 2\mathrm{H}_{2} \mathrm{O} \)

With this balanced equation, you can correctly predict the amounts of reactants and products involved, making it pivotal for any stoichiometry calculations.

Molar Mass Calculation

Molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). It is a fundamental concept used to convert between mass and moles, crucial for stoichiometry.
For example, to find the molar mass of \( \mathrm{NH}_4 \mathrm{Cl} \):

• Sum up the molar masses of each element in the compound: nitrogen (N), hydrogen (H), and chlorine (Cl).
• \( \mathrm{NH}_4 \mathrm{Cl} \) has a molar mass of 53.49 g/mol obtained by adding the atomic masses:\( 14.01 \ ext{(N)} + 4 \times 1.01 \ ext{(H)} + 35.45 \ ext{(Cl)} \).

Accurate molar mass calculations allow us to convert given amounts from grams to moles efficiently. This conversion is necessary to further calculate other values in a chemical reaction, such as identifying the limiting reactant or determining the amount of products formed. Remember, this step is indispensable for any stoichiometry problem-solving.