Question

What is the mass of the solid \(\mathrm{NH}_{4} \mathrm{Cl}\) formed when \(73.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) is mixed with an equal mass of \(\mathrm{HCl} ?\) What is the volume of the gas remaining, measured at \(14.0^{\circ} \mathrm{C}\) and \(752 \mathrm{mmHg}\) ? What gas is it?

Short answer

107.0 g of \(\text{NH}_4\text{Cl}\) is formed, and 54.3 L of \(\text{NH}_3\) gas remains.

Step-by-step solution

Write the Chemical Equation

The chemical reaction between ammonia \(\text{NH}_3\) and hydrochloric acid \(\text{HCl}\) to form ammonium chloride \(\text{NH}_4\text{Cl}\) is: \[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl} \] This shows that one mole of \(\text{NH}_3\) reacts with one mole of \(\text{HCl}\) to form one mole of \(\text{NH}_4\text{Cl}\).

Calculate Moles of Reactants

Given \(73.0\, \text{g}\) each of \(\text{NH}_3\) and \(\text{HCl}\), we need to find the number of moles: \[ \text{Molar mass of NH}_3 = 14.01 + 3\times 1.01 = 17.04\, \text{g/mol} \] \[ \text{Moles of NH}_3 = \frac{73.0}{17.04} \approx 4.28\, \text{mol} \] \[ \text{Molar mass of HCl} = 1.01 + 35.45 = 36.46\, \text{g/mol} \] \[ \text{Moles of HCl} = \frac{73.0}{36.46} \approx 2.00\, \text{mol} \]

Determine Limiting Reactant

The reaction is a 1:1 ratio, so comparing the moles: \(4.28\, \text{mol of } \text{NH}_3\) and \(2.00\, \text{mol of } \text{HCl}\), \(\text{HCl}\) is the limiting reactant because it has fewer moles.

Calculate Mass of \(\text{NH}_4\text{Cl}\) Formed

Since \(\text{HCl}\) is the limiting reactant, it will completely react to form \(2.00\, \text{mol}\) of \(\text{NH}_4\text{Cl}\). The molar mass of \(\text{NH}_4\text{Cl}\) is \(14.01 + 4\times1.01 + 35.45 = 53.50\, \text{g/mol}\). The mass formed is: \[ \text{Mass of NH}_4\text{Cl} = 2.00\, \text{mol} \times 53.50\, \text{g/mol} = 107.0\, \text{g} \]

Determine Remaining Gas

Since \(\text{HCl}\) is the limiting reactant, all \(2.00\, \text{mol of HCl}\) will react, leaving \(4.28 - 2.00 = 2.28\, \text{mol of } \text{NH}_3\) remaining unreacted.

Calculate Volume of Remaining Gas

Use the ideal gas law \(PV = nRT\) to find the volume. Convert \(14.0^{\circ}\text{C}\) to Kelvin: \[ T = 14.0 + 273.15 = 287.15\, \text{K} \] Convert pressure to atmospheres: \[ P = \frac{752}{760} = 0.989\, \text{atm} \] Use \(R = 0.0821\, \text{L atm/mol K}\), and substitute: \[ V = \frac{nRT}{P} = \frac{2.28 \times 0.0821 \times 287.15}{0.989} \approx 54.3\, \text{L} \]

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that determines the maximum amount of product that can be formed. It is called the limiting reactant because it limits the chemical reaction from progressing when it is completely consumed. When you know the limiting reactant, you can calculate the theoretical yield of the reaction, which is the maximum possible mass of product that can be obtained.

In the given example, we have two reactants: ammonia (\( \text{NH}_3 \)) and hydrochloric acid (\( \text{HCl} \)). The chemical equation shows a 1:1 mole ratio, which means that for each mole of \( \text{NH}_3 \), one mole of \( \text{HCl} \) is needed. When you calculate the moles of each, \( \text{HCl} \) has fewer moles (2.00 moles compared to 4.28 moles of \( \text{NH}_3\)), making \( \text{HCl} \) the limiting reactant. Once all the \( \text{HCl} \) is used, the reaction stops, and the remaining \( \text{NH}_3 \) cannot further react.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry, represented as \( PV = nRT \). This relationship connects pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), the gas constant (\(R\)), and temperature (\(T\)) to describe the state of a gas under ideal conditions.

• \(P\) is the pressure of the gas.
• \(V\) is the volume occupied by the gas.
• \(n\) is the number of moles of gas.
• \(R\) is the ideal gas constant (0.0821 L atm / mol K).
• \(T\) is the temperature in Kelvin.

To calculate the volume of the remaining gas after a chemical reaction, you rearrange the equation to \( V = \frac{nRT}{P} \). Using the values of \( n = 2.28 \) moles, \( T = 287.15 \) K, and \( P = 0.989 \) atm, you can find the volume. The calculations show that the volume of remaining \( \text{NH}_3 \) gas is roughly 54.3 liters.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products, characterized by the breaking and forming of chemical bonds. Each reaction is represented by a chemical equation that shows how the reactants are converted to products with a specific stoichiometry, which means the exact balance of atoms and molecules.

In this particular reaction, ammonia (\( \text{NH}_3 \)) reacts with hydrochloric acid (\( \text{HCl} \)) to form ammonium chloride (\( \text{NH}_4\text{Cl} \)). The balanced chemical equation is:\[ \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl} \].
This equation demonstrates that one mole of ammonia reacts with one mole of hydrochloric acid to produce one mole of ammonium chloride. Stoichiometry gives us a precise ratio to follow when calculating the amounts of products formed and reactants consumed.

Molar Mass

The concept of molar mass is essential in stoichiometry, allowing us to convert between the mass of a substance and the number of moles. The molar mass is the mass of one mole of a substance, expressed in grams/mol, and it is calculated by adding up the atomic masses of all the atoms in a chemical formula.

For example, the molar mass of ammonia (\( \text{NH}_3 \)) is calculated by adding the atomic masses of one nitrogen atom and three hydrogen atoms: \( 14.01 \, \text{g/mol} + 3 \times 1.01 \, \text{g/mol} = 17.04 \, \text{g/mol} \). Similarly, the molar mass of \( \text{HCl} \) is \( 36.46 \, \text{g/mol} \), and for \( \text{NH}_4\text{Cl} \), it is \( 53.50 \, \text{g/mol} \).

Understanding molar mass helps to determine the amounts of substances involved in a chemical reaction. It is crucial when you need to calculate how much product is formed from a given quantity of reactants.