Problem 113 - Chemistry The Science in Context

Question

Ammonia rapidly reacts with hydrogen chloride, making ammonium chloride. Write a balanced chemical equation for the reaction, and calculate the number of grams of excess reactant when \(3.0 \mathrm{g}\) of \(\mathrm{NH}_{3}\) reacts with \(5.0 \mathrm{g}\) of \(\mathrm{HCl}\).

Step-by-Step Solution

Answer

Answer: There are 0.66 grams of excess NH3 remaining in this reaction.

Step 1: Write the balanced chemical equation

The first step is to write the balanced chemical equation for the reaction. When ammonia reacts with hydrogen chloride, it forms ammonium chloride: \[ \mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4}\mathrm{Cl}(s) \] This equation is already balanced, with one mole of each reactant producing one mole of product.

Step 2: Calculate the moles of both reactants

Next, we need to convert the masses of reactants given in grams to moles. To do this, we divide the mass of each reactant by its molar mass. The molar mass of NH3 is 14.01 g/mol (for nitrogen) + 3(1.01 g/mol) (for hydrogen) = 17.03 g/mol, and the molar mass of HCl is 1.01 g/mol (for hydrogen) + 35.45 g/mol (for chlorine) = 36.46 g/mol. Moles of NH3 = \(\frac{3.0 \ \mathrm{g}}{17.03 \ \mathrm{g/mol}} = 0.176 \ \mathrm{mol}\) Moles of HCl = \(\frac{5.0 \ \mathrm{g}}{36.46 \ \mathrm{g/mol}} = 0.137 \ \mathrm{mol}\)

Step 3: Determine the limiting reactant

To determine the limiting reactant, we need to compare the mole ratio of both reactants. In this reaction, the stoichiometry is 1:1, which means one mole of NH3 reacts with one mole of HCl to produce one mole of NH4Cl. Since there are 0.176 mol of NH3 and 0.137 mol of HCl, it is clear that HCl is the limiting reactant because there is a smaller amount of it available to react.

Step 4: Calculate the grams of excess reactant

Now that we know HCl is the limiting reactant, we can use its moles to find the moles of NH3 that reacted: Moles of NH3 reacted = 0.137 mol (since it's a 1:1 ratio) Excess moles of NH3 = Initial moles of NH3 - Moles of NH3 reacted = 0.176 mol - 0.137 mol = 0.039 mol Now we can convert the excess moles of NH3 back to grams using the molar mass of NH3: Grams of excess NH3 = 0.039 mol x 17.03 g/mol = 0.66 g So, there are 0.66 grams of excess NH3 in this reaction.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the calculations of reactants and products in chemical reactions. It involves using the balanced chemical equation to determine the relative amounts of substances involved. By understanding stoichiometry, you can predict how much product can be formed from given reactants or how much of a reactant is needed for a complete reaction.

• First, write the balanced chemical equation. This is crucial because it provides the exact proportion of each element involved in the reactions.
• The coefficients in a balanced equation represent the ratio of moles of each substance.
• Use these ratios to calculate the moles of reactants required or the moles of products formed.

Overall, stoichiometry is like a recipe in cooking: follow the right proportions to ensure everything comes out perfectly! It guides you on how much of each ingredient to use and what amounts you'll end up with.

Limiting Reactant

The concept of the limiting reactant is central to understanding reactions fully. It determines how much product can ultimately be formed in a chemical reaction. The limiting reactant is the substance that is completely consumed first, stopping the reaction from continuing to produce more product.

• Identify the limiting reactant by comparing the mole ratio from the balanced chemical equation with the actual moles you have.
• Once the limiting reactant is used up, the reaction cannot proceed, making it crucial to identify it correctly.

In our exercise, the limiting reactant was HCl. We determined this because, despite having more moles of ammonia, the moles of HCl were less, hence it limits the amount of product formed.

Molar Mass

Molar mass is the mass of one mole of a given substance (usually in grams/mole) and is key to converting between grams and moles. It’s defined as the sum of the atomic masses of all atoms in a molecule.

• To calculate molar mass, simply add the atomic masses from the periodic table for all atoms in a molecule.
• For example, ammonia \(NH_3\) has a molar mass of 17.03 g/mol: \(14.01 \, \text{g/mol for Nitrogen} + 3 \times 1.01 \, \text{g/mol for Hydrogen}\).

Knowing the molar mass allows us to convert grams to moles by dividing the mass in grams by the molar mass. This step is crucial in stoichiometric calculations.

Balanced Chemical Equation

A balanced chemical equation is foundational in chemistry as it reflects the Law of Conservation of Mass, which states that mass cannot be created or destroyed. A balanced equation correctly shows the number of atoms of each element that are conserved in a reaction.

• In the exercise, the equation is \( ext{NH}_3(g) + ext{HCl}(g) \rightarrow ext{NH}_4 ext{Cl}(s)\).
• Each side of the equation has the same number of each type of atom, fulfilling \(1:1:1\) ratio for nitrogen, hydrogen, and chlorine.

Ensuring balance in an equation is critical for accurately performing stoichiometric calculations, as it validates the quantities you are working with.