Question

Small quantities of ammonia gas can be generated in the laboratory by heating an ammonium salt with a strong base. For example, ammonium chloride reacts with sodium hydroxide according to the following balanced equation: $$ \mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{NaOH}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ What mass of ammonia gas is produced if \(1.39 \mathrm{~g}\) of ammonium chloride reacts completely?

Short answer

The mass of ammonia gas produced when \(1.39 \mathrm{~g}\) of ammonium chloride reacts completely is approximately \(0.4421 \mathrm{~g}\).

Step-by-step solution

Write down the given information

We are given the balanced chemical equation: \[ \mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{NaOH}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g) \] We are also given the mass of ammonium chloride: \(1.39 \mathrm{~g}\)

Calculate the moles of ammonium chloride

To convert the mass of ammonium chloride to moles, we will use the molar mass of ammonium chloride. The molar mass of ammonium chloride \(\mathrm{NH}_{4} \mathrm{Cl}\) is: \(14.01 \mathrm{~g/mol}\) (for N) + \(1.01 \mathrm{~g/mol} \times 4\) (for 4 H) + \(35.45 \mathrm{~g/mol}\) (for Cl) = \(53.49 \mathrm{~g/mol}\) Now we can calculate the moles of ammonium chloride: \[ \text{moles of } \mathrm{NH}_{4} \mathrm{Cl} = \frac{\text{mass of } \mathrm{NH}_{4} \mathrm{Cl}}{\text{molar mass of } \mathrm{NH}_{4} \mathrm{Cl}} = \frac{1.39 \mathrm{~g}}{53.49 \mathrm{~g/mol}} = 0.02598 \mathrm{~mol} \]

Use the stoichiometry of the balanced equation to find moles of ammonia produced

According to the balanced chemical equation, 1 mole of ammonium chloride reacts to produce 1 mole of ammonia gas. Therefore, the moles of ammonia produced are equal to the moles of ammonium chloride: \[ \text{moles of }\mathrm{NH}_{3} =\text{ moles of }\mathrm{NH}_{4} \mathrm{Cl} = 0.02598 \mathrm{~mol} \]

Convert moles of ammonia to mass

To find the mass of ammonia gas produced, we will use the molar mass of ammonia, \(\mathrm{NH}_{3}\). The molar mass of ammonia is: \(14.01 \mathrm{~g/mol}\) (for N) + \(1.01 \mathrm{~g/mol} \times 3\) (for 3 H) = \(17.03 \mathrm{~g/mol}\) Now we can calculate the mass of ammonia produced: \[ \text{mass of } \mathrm{NH}_{3} = \text{moles of } \mathrm{NH}_{3} \times \text{molar mass of } \mathrm{NH}_{3} = 0.02598 \mathrm{~mol} \times 17.03 \mathrm{~g/mol} = 0.4421 \mathrm{~g} \]

State the final answer

The mass of ammonia gas produced when \(1.39 \mathrm{~g}\) of ammonium chloride reacts completely is approximately \(0.4421 \mathrm{~g}\).

Key concepts

Chemical Reactions

In chemistry, a chemical reaction involves transforming one or more substances into different substances. It's like a chemical "exchange" where old bonds break, and new ones form. This process is not just a mix-up but a rearrangement of the atoms. During a chemical reaction, the substances you start with are called reactants, and the new substances formed are products.
For example, in the original exercise, ammonium chloride (\(\mathrm{NH}_{4} \mathrm{Cl}\)) and sodium hydroxide (\(\mathrm{NaOH}\)) come together to form new products: ammonia gas (\(\mathrm{NH}_{3}\)), sodium chloride (\(\mathrm{NaCl}\)), and water (\(\mathrm{H}_{2} \mathrm{O}\)).
This transformation is defined by a chemical equation, where the arrow shows the direction of the reaction:

• Reactants are listed on the left side.
• Products are listed on the right side.

Recognizing chemical reactions is crucial in chemistry as they underlie many processes in nature and industry.

Molar Mass Calculation

Molar mass is like a chemical's "weight" in grams per mole and is essential for converting between the weight of a substance and the amount of substance in moles. It's particularly important in stoichiometry because it links a substance's mass to the quantities used in chemical reactions.
To calculate the molar mass, sum the atomic masses (in grams per mole) of all the atoms in a molecule. For ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\), it's calculated as follows:

• Nitrogen (N): 14.01 g/mol
• Hydrogen (H): 1.01 g/mol each, so 1.01 g/mol * 4 = 4.04 g/mol
• Chlorine (Cl): 35.45 g/mol

The total molar mass is 14.01 + 4.04 + 35.45 = 53.49 g/mol. Using this, you can convert grams to moles, which is critical for understanding how much of a substance will react or be produced according to a chemical equation.

Balanced Chemical Equation

A balanced chemical equation is like a balanced diet—it ensures everything in the reaction is accounted for. This is crucial in chemistry to satisfy the Law of Conservation of Mass, which states that no atoms are lost or gained in a chemical reaction.
To balance an equation, count the number of each atom on the reactant side and adjust the coefficients (the numbers in front) to make them equal the product side. For example:

• In the equation given in the exercise: \(\mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{NaOH}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g) \), every type of atom is evenly distributed between reactants and products.
• There's one nitrogen, four hydrogens from ammonium, one sodium, one chlorine, and the combined atoms of water on each side.

This systematic approach means you know exactly how much of each substance participates in a reaction, which is key to predicting the quantities of products formed. Always remember to balance the equation before using it in calculations, like finding out the amount of ammonia produced!