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 -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?

First, we need to determine which gas is the limiting reactant. We are given the mass of both gases and can use their molar mass to convert to moles. Moles of \(NH_3\): \(\frac{5.00 \mathrm{g}}{17.03 \mathrm{g/mol}}=0.2937 \mathrm{mol}\) Moles of \(HCl\): \(\frac{5.00 \mathrm{g}}{36.46 \mathrm{g/mol}}=0.1372 \mathrm{mol}\) The stoichiometry of the balanced equation is 1:1. Since there are more moles of \(NH_3\) than \(HCl\), \(HCl\) is the limiting reactant.

Now, we need to determine the amount of each reactant left after the reaction is complete. Since \(HCl\) will be completely consumed, there will be no moles of \(HCl\) left. The moles of \(NH_3\) left can be calculated. Moles of \(NH_3\) left: \(0.2937 - 0.1372 = 0.1565 \mathrm{mol}\)

The initial pressure of the system can be calculated using the ideal gas law, \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin. The temperature is given as \(25^{\circ} \mathrm{C}\), which is equal to \(298.15 \mathrm{K}\). The ideal gas constant, \(R\), is \(0.0821 \mathrm{L \cdot atm/mol \cdot K}\). Initially, only \(NH_3\) is in the first flask, so: Initial pressure (both flasks): \(P_1=\frac{n_1 \cdot R \cdot T_1}{V_1}\) \(P_1=\frac{0.2937 \mathrm{mol} \cdot 0.0821 \mathrm{L \cdot atm/mol \cdot K} \cdot 298.15 \mathrm{K}}{2.00 \mathrm{L}}\) \(P_1=3.63 \mathrm{atm}\) After the reaction, the final pressure can be calculated using the moles of \(NH_3\) left: Final pressure (both flasks): \(P_2=\frac{n_2 \cdot R \cdot T_2}{V_2}\) \(P_2=\frac{0.1565 \mathrm{mol} \cdot 0.0821 \mathrm{L \cdot atm/mol \cdot K} \cdot 298.15 \mathrm{K}}{4.00 \mathrm{L}}\) \(P_2=0.968 \mathrm{atm}\)

Finally, we need to determine the mass of ammonium chloride formed from the reaction. Since \(HCl\) is the limiting reactant, the mass of ammonium chloride produced can be calculated using the stoichiometry of the balanced equation. Moles of \(NH_4Cl\) produced: \(0.1372 \mathrm{mol}\) Mass of \(NH_4Cl\) produced: \(0.1372 \mathrm{mol} \times 53.49 \mathrm{g/mol} = 7.34 \mathrm{g}\) In summary: a) The gas that will remain in the system after the reaction is complete is ammonia (\(NH_3\)). b) The final pressure of the system after the reaction is complete is 0.968 atm. c) The mass of ammonium chloride formed is 7.34 g.