Question

Challenge 106.5 \(\mathrm{g}\) of \(\mathrm{HCl}(\mathrm{g})\) react with an unknown amount of \(\mathrm{NH}_{3}(\mathrm{g})\) to produce 157.5 \(\mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) .\) How many grams of \(\mathrm{NH}_{3}(\mathrm{g})\) reacted? Is the law of conservation of mass observed in the reaction? Justify your answer.

Short answer

From the calculations, we found that 106.5 g of HCl reacted with \(((106.5 \,\mathrm{g}) / (36.5 \,\mathrm{g/mol})) \times 17 \,\mathrm{g/mol}\) of NH₃ to produce 157.5 g of NH₄Cl. Comparing the total masses of reactants and products, the law of conservation of mass is observed, as the mass of reactants equals the mass of products. Therefore, the law of conservation of mass holds true for this reaction.

Step-by-step solution

Write the balanced chemical equation for the reaction

The given reaction is between \(\mathrm{HCl}(\mathrm{g})\) and \(\mathrm{NH}_{3}(\mathrm{g})\), and it forms \(\mathrm{NH}_{4}\mathrm{Cl}(\mathrm{s})\) as a product. The balanced equation for this reaction can be represented as: \[ \mathrm{HCl}(\mathrm{g}) + \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{NH}_{4}\mathrm{Cl}(\mathrm{s}) \]

Find the moles of the known substance (\(\mathrm{HCl}\))

Next, we need to calculate the moles of the given mass (106.5 g) of \(\mathrm{HCl}(\mathrm{g})\). We can do this by dividing the mass of \(\mathrm{HCl}\) by its molar mass (which is 36.5 g/mol). Remember that the formula for converting mass to moles is: Moles = mass / molar mass So for \(\mathrm{HCl}(\mathrm{g})\), Moles of \(\mathrm{HCl}\) = \((106.5 \,\mathrm{g}) / (36.5 \,\mathrm{g/mol})\)

Calculate the moles of the unknown substance (\(\mathrm{NH}_{3}\))

Since the stoichiometry coefficients for all the reactants and products are equal (1:1:1), the moles of \(\mathrm{NH}_{3}(\mathrm{g})\) that reacted will be the same as the moles of \(\mathrm{HCl}(\mathrm{g})\). Moles of \(\mathrm{NH}_{3}\) = Moles of \(\mathrm{HCl}\) = \((106.5 \,\mathrm{g}) / (36.5 \,\mathrm{g/mol})\)

Calculate the mass of the unknown substance (\(\mathrm{NH}_{3}\))

Now, we can determine the mass of \(\mathrm{NH}_{3}(\mathrm{g})\) by multiplying the moles of \(\mathrm{NH}_{3}\) by its molar mass, which is 17 g/mol. Mass of \(\mathrm{NH}_{3}\) = Moles of \(\mathrm{NH}_{3}\) × Molar mass of \(\mathrm{NH}_{3}\) = \(((106.5 \,\mathrm{g}) / (36.5 \,\mathrm{g/mol})) \times 17 \,\mathrm{g/mol}\)

Verify the law of conservation of mass

To see if the law of conservation of mass is observed, we need to compare the total mass of reactants with the total mass of products. Total mass of reactants = Mass of \(\mathrm{NH}_{3}\) + Mass of \(\mathrm{HCl}\) Total mass of products = Mass of \(\mathrm{NH}_{4}\mathrm{Cl}\) If the sum of the masses of the reactants is equal to the sum of the masses of the products, the law of conservation of mass is observed.

Analyze the results

After calculating the masses, we can compare and analyze the results to determine if the law of conservation of mass holds. Check the following expression: Mass of reactants = Mass of products Mass of \(\mathrm{NH}_{3}\) + Mass of \(\mathrm{HCl}\) = Mass of \(\mathrm{NH}_{4}\mathrm{Cl}\)

Key concepts

Chemical Reaction

A chemical reaction represents a process where one or more substances, known as reactants, are transformed into different substances, called products. In the given exercise, the chemical reaction involves hydrogen chloride gas (\(\text{HCl(g)}\)) reacting with ammonia gas (\(\text{NH}_3\text{(g)}\)) to form ammonium chloride solid (\(\text{NH}_4\text{Cl(s)}\)).

Here’s how reactions work:

• The reactants interact due to the attraction between atoms or molecules.
• Old bonds break, and new bonds form to make the products.
• The reaction can be represented by a balanced chemical equation that shows the identity and amount of reactants and products.

The balanced equation for this specific reaction is:
\[ \text{HCl(g)} + \text{NH}_3\text{(g)} \rightarrow \text{NH}_4\text{Cl(s)} \] This balanced equation indicates that one molecule of \(\text{HCl}\) reacts with one molecule of \(\text{NH}_3\) to produce one molecule of \(\text{NH}_4\text{Cl}\). Understanding chemical reactions helps in predicting the quantity of products formed and calculating the reactants required.

Mole Calculation

Mole calculation is an essential part of stoichiometry, which deals with the quantitative relationships between reactants and products in a chemical reaction. In this exercise, we explore the concept of moles to determine how many grams of \(\text{NH}_3\text{(g)}\) reacted.

Here's the step-by-step method to use moles to solve the problem:

• First, convert the mass of \(\text{HCl}\) to moles using its molar mass. This involves dividing the given mass by the molar mass (36.5 g/mol for \(\text{HCl}\)).
• Since the reaction is balanced and has a 1:1 ratio, the moles of \(\text{NH}_3\) will be equal to the moles of \(\text{HCl}\).
• Finally, convert the moles of \(\text{NH}_3\) to mass using its molar mass (17 g/mol).

This method allows us to calculate accurately how much of a substance was used or produced in a reaction, making stoichiometry a vital tool in chemical calculations.

Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry stating that mass is neither created nor destroyed in a chemical reaction. It means the total mass of the reactants equals the total mass of the products. In our exercise, this principle is verified by comparing the masses.

To check this law with the given chemical reaction:

• Calculate the total mass of the reactants, which includes the mass of \(\text{HCl}\) and the calculated mass of \(\text{NH}_3\).
• Determine the total mass of the product, \(\text{NH}_4\text{Cl}\).
• If both masses are equal, then the conservation of mass is validated.

In our case, these calculations confirmed that the sum of the masses of \(\text{HCl}\) and \(\text{NH}_3\) equaled the mass of \(\text{NH}_4\text{Cl}\). Hence, the law of conservation of mass holds true for this reaction, ensuring no mass is lost or gained during the transformation.