Problem 122 - General Chemistry: Principles and Modern Applications

Question

For a specific reaction, ammonium dichromate is the only reactant and chromium(III) oxide and water are two of the three products. What is the third product and how many grams of this product are produced per kilogram of ammonium dichromate decomposed?

Step-by-Step Solution

Answer

The third product is Nitrogen gas (\( N_2 \)). Approximately 111.17 g of this product are produced per kilogram of ammonium dichromate decomposed.

Step 1: Identify the Balanced Chemical Reaction

Firstly, the balanced chemical reaction for the decomposition of ammonium dichromate (\( (NH_4)_2Cr_2O_7 \)) is: \[ (NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + 4 H_2O + N_2 \] So, the third product is Nitrogen gas (\( N_2 \))

Step 2: Calculate Molar Masses

Next, calculate the molar masses of the reactant and the third product: \[ Molar\ mass\ of\ (NH_4)_2Cr_2O_7 = 252.07\ g/mol \] \[ Molar\ mass\ of\ N_2 = 28.014\ g/mol \]

Step 3: Calculate the Amount of Third Product Produced

The stoichiometry of the reaction tells us that 1 mole of \( (NH_4)_2Cr_2O_7 \) decomposed yields 1 mole of \( N_2 \). So, if we have 1000 g (1 kg) of \( (NH_4)_2Cr_2O_7 \), this corresponds to \( \frac{1000}{252.07} \) moles. Hence, the same number of moles of \( N_2 \) is produced, which corresponds to \( \frac{1000}{252.07} \times 28.014 = 111.17 \) g of \( N_2 \).

Key Concepts

Balanced Chemical Equation

In a chemical reaction, it's crucial to have a balanced chemical equation. This ensures the law of conservation of mass is satisfied, which states that mass cannot be created or destroyed in a chemical reaction. All atoms present in the reactants must be accounted for in the products. In our example, the decomposition of ammonium dichromate is balanced as follows: \[(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + 4 H_2O + N_2\] Examining this equation, you can see that the atoms on both sides are equal. Each nitrogen, hydrogen, chromium, and oxygen atom in the reactant matches with the respective atoms in the products. The third product formed is nitrogen gas \(N_2\). To determine this, you must balance and count the elements of each product to ensure the total count remains the same.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It uses the coefficients from the balanced equation to make calculations about the masses of reactants and products. Let's apply stoichiometry to our decomposition reaction: For every mole of \((NH_4)_2Cr_2O_7\) that decomposes, one mole of each product is formed, including nitrogen gas \(N_2\). This mole-to-mole ratio is crucial for calculating how much of a substance is produced or used up in a reaction. By using these ratios derived from the balanced equation, you can determine quantities involved in the reactions, such as in our case, calculating the grams of nitrogen gas produced from a certain amount of ammonium dichromate.

Molar Mass Calculation

Molar mass is a fundamental concept for connecting the mass of a substance to the amount in moles. It is the mass of one mole of a given substance (atoms, molecules, etc.) and is usually expressed in grams per mole \(g/mol\). Calculating molar mass correctly enables precise stoichiometric calculations. For the decomposition reaction of ammonium dichromate:

• The molar mass of \((NH_4)_2Cr_2O_7\) is calculated by adding the atomic masses of all its constituent atoms: \(252.07\ g/mol\).
• The molar mass of nitrogen gas \(N_2\) is the sum of the masses of two nitrogen atoms: \(28.014\ g/mol\).

By determining these molar masses, you can transform your stoichiometric calculations from moles to grams, which are more practical for real-world applications. This allows you to conclude that 1000 grams of ammonium dichromate (H_2O\ldots) will produce 111.17 grams of nitrogen gas by using these calculations and the earlier computed stoichiometric ratios.