Ammonia and hydrogen chloride react to form solid ammonium chloride: $$\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$$ Two \(2.00-\mathrm{L}\) flasks at \(25^{\circ} \mathrm{C}\) are connected by a valve, as shown in the drawing. One flask contains \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g),\) and the other contains \(5.00 \mathrm{~g}\) of \(\mathrm{HCl}(g) .\) When the valve is opened, the gases react until one is completely consumed. (a) Which gas will remain in the system after the reaction is complete? (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.) (c) What mass of ammonium chloride will be formed?

Step by step solution

01

Calculate the Number of Moles

First, we need to calculate the number of moles of NH3 and HCl. The molar mass of NH3 is 14.01 (N) + 3 × 1.01 (H) = 17.04 g/mol and that of HCl is 1.01 (H) + 35.45 (Cl) = 36.46 g/mol. Using the given masses and molar masses, we can calculate the number of moles. For NH3: Number of moles of NH3 = \( \frac{mass}{molar \thinspace mass} \) Number of moles of NH3 = \( \frac{5.00 \thinspace g}{17.04 \thinspace g/mol} \) = 0.293 mol For HCl: Number of moles of HCl = \( \frac{mass}{molar \thinspace mass} \) Number of moles of HCl = \( \frac{5.00 \thinspace g}{36.46 \thinspace g/mol} \) = 0.137 mol

02

Compare the mole ratios and find which gas is limiting

We can compare the mole ratios of NH3 and HCl, and determine the limiting reagent, i.e., the reagent which will be completely consumed first. The required mole ratio for the reaction is 1:1. So, we can divide the number of moles of each gas by their respective required moles. For NH3: \( \frac{0.293 \thinspace mol}{1} \) = 0.293 For HCl: \( \frac{0.137 \thinspace mol}{1} \) = 0.137 Since 0.137 < 0.293, HCl is the limiting reagent and will be completely consumed.

03

Calculate the number of moles of the remaining gas and the moles of the product formed

We know that HCl is the limiting reagent, so we will calculate the remaining moles of NH3 and the moles of the product, NH4Cl. The remaining moles of NH3 = Initial moles of NH3 - moles of HCl reacted The remaining moles of NH3 = 0.293 - 0.137 = 0.156 mol The moles of NH4Cl formed = moles of HCl reacted = 0.137 mol

04

Calculate the final total pressure of the system

Now, we will calculate the final pressure of the system using the Ideal Gas Law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We will first convert the given temperature from Celsius to Kelvin. T = 25 + 273.15 = 298.15 K To find the final pressure (P), we will use the remaining moles of NH3 (n), the volume (V = 2 L + 2 L = 4 L), the gas constant (R = 0.0821 L atm/mol K), and the temperature (T). P = \( \frac{nRT}{V} \) P = \( \frac{0.156 \thinspace mol \times 0.0821 \thinspace L \thinspace atm \thinspace / \thinspace mol \thinspace K \times 298.15 \thinspace K}{4 \thinspace L} \) = 0.959 atm

05

Calculate the mass of ammonium chloride formed

Finally, we will calculate the mass of NH4Cl formed using the moles of NH4Cl and its molar mass. The molar mass of NH4Cl is 14.01 (N) + 4 × 1.01 (H) + 35.45 (Cl) = 53.50 g/mol. mass of NH4Cl = moles × molar mass mass of NH4Cl = 0.137 mol × 53.50 g/mol = 7.33 g #Answers#: (a) NH3 gas will remain in the system after the reaction is complete. (b) The final pressure of the system will be 0.959 atm. (c) The mass of ammonium chloride formed will be 7.33 g.

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