Question

Ammonia and gaseous hydrogen chloride combine to form ammonium chloride. $$ \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s) $$ If \(4.21 \mathrm{L}\) of \(\mathrm{NH}_{3}(g)\) at \(27^{\circ} \mathrm{C}\) and 1.02 atm is combined with 5.35 L of \(\mathrm{HCl}(g)\) at \(26^{\circ} \mathrm{C}\) and 0.998 atm, what mass of \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) will be produced? Which gas is the limiting reactant? Which gas is present in excess?

Short answer

In conclusion, 9.04 grams of ammonium chloride (NH4Cl) will be produced from the given amounts of reactants. The limiting reactant is ammonia (NH3), and the excess reactant is hydrogen chloride (HCl).

Step-by-step solution

Convert the given volume, temperature, and pressure for each gas to moles.

Using the Ideal Gas Law equation, n = PV/RT, we will calculate the moles of ammonia (NH3) and hydrogen chloride (HCl). For ammonia, NH3: \(P = 1.02\; atm\) \(V = 4.21\; L\) \(T = (27°C + 273) = 300\; K\) \(R = 0.0821 \;L * atm /(K * mol)\) \(n_{NH_3} = \frac{P * V}{R * T} = \frac{1.02 * 4.21}{0.0821 * 300} = 0.169\; moles\) For hydrogen chloride, HCl: \(P = 0.998 \;atm\) \(V = 5.35\; L\) \(T = (26°C + 273) = 299\; K\) \(R = 0.0821\; L * atm /(K * mol)\) \(n_{HCl} = \frac{P * V}{R * T} = \frac{0.998 * 5.35}{0.0821 * 299} = 0.228\; moles\)

Determine the limiting and excess reactants.

Given the stoichiometry of the balanced equation, we know that the mole ratio of NH3 to HCl is 1:1. So first, we will calculate the mole ratio between NH3 and HCl present in the given amounts. Mole ratio \(= \frac{n_{NH_3}}{n_{HCl}} = \frac{0.169}{0.228} = 0.741\) Since the mole ratio = 0.741 < 1; That means, NH3 will be used up first and there will be some HCl left after reaction completion. Hence, NH3 is the limiting reactant and HCl is the excess reactant.

Calculate the mass of ammonium chloride formed.

To find the mass of ammonium chloride formed, we will use the stoichiometry of the balanced equation and the number of moles of the limiting reactant NH3. For every mole of NH3 used, 1 mole of NH4Cl is formed. Therefore, moles of NH4Cl formed = moles of NH3 used Now we need to convert moles to mass using the molecular mass of NH4Cl: Molecular mass of NH4Cl = 14.01 (N) + 4 * 1.01 (H) + 35.45 (Cl) = 53.49 g/mol Finally, we calculate the mass formed: Mass of NH4Cl formed = moles of NH4Cl formed * molecular mass of NH4Cl Mass of NH4Cl formed \(= 0.169\; moles * 53.49\; g/mol = 9.04\; g\) In conclusion, 9.04 grams of ammonium chloride (NH4Cl) will be produced from the given amounts of reactants. The limiting reactant is ammonia (NH3), and the excess reactant is hydrogen chloride (HCl).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, temperature, and amount (in moles) of a gas. The equation is expressed as \( PV = nRT \), where:

• \( P \) represents the pressure of the gas
• \( V \) is the volume
• \( n \) stands for the number of moles
• \( R \) is the ideal gas constant, usually 0.0821 L atm/mol K
• \( T \) is the temperature in Kelvin

To use the Ideal Gas Law effectively, it's crucial to convert all measurements to their correct units, particularly temperature to Kelvin and volume to liters, if necessary. By applying the Ideal Gas Law to the given problem, we can determine the number of moles of ammonia \((\text{NH}_3)\) and hydrogen chloride \((\text{HCl})\) by rearranging the formula to \( n = \frac{{PV}}{{RT}} \). This conversion helps us in subsequent stoichiometric calculations.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely consumed first, thereby limiting the extent of the reaction and determining the maximum amount of product that can be formed. To identify the limiting reactant, we compare the mole ratio of the reactants present to the ratio required by the balanced chemical equation.In the given chemical reaction \(( \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) )\), the ideal ratio is 1:1. By calculating the moles of each reactant (as we did with the Ideal Gas Law), and comparing them to the stoichiometric ratio, we find that \( \text{NH}_3 \) is present in a lesser amount relative to \( \text{HCl} \). Therefore, ammonia \((\text{NH}_3)\) is the limiting reactant. Knowing which reactant is limiting is essential, as it directly impacts how much product can be expected.

Mole Calculations

Mole calculations are a core part of stoichiometry, allowing us to convert between the amount of substance and its mass. Understanding moles helps bridge the gap between microscopic subatomic world and macroscopic laboratory experiments.Using the Ideal Gas Law, we first found the moles of \( \text{NH}_3 \) and \( \text{HCl} \). The equation \( n = \frac{{PV}}{{RT}} \) leverages measured conditions of a gas to calculate its quantity in moles. Once we identify the limiting reactant, we use its amount to calculate the amount of product formed.In this particular reaction, 0.169 moles of \(\text{NH}_3\) directly translate into 0.169 moles of \(\text{NH}_4\text{Cl}\), because they react in a 1:1 mole ratio according to the balanced chemical equation. Mastery over mole conversions is key to evaluating chemical outcomes.

Chemical Reactions

A chemical reaction transforms reactants into products through the rearrangement of atoms. The reaction in question is synthesis: \( \text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s) \). In this process, gaseous ammonia and hydrogen chloride combine to form a solid, ammonium chloride.The balanced equation provides a mole ratio that guides stoichiometric calculations—another critical reason for balancing chemical equations effectively. This equation helps us determine theoretical yields and identify the limiting reactant, which dictates the maximum potential formation of product.By understanding the conditions and quantities involved, students can further conclude the mass of the product, using molar masses. For \( \text{NH}_4\text{Cl} \), the molar mass is calculated as 53.49 g/mol. Converting moles to grams illustrates a complete cycle from initial conditions to experimental data.