Step-by-step explanation:
To solve this problem, we first need to calculate the number of moles of NH3 and HCl using the ideal gas law:
n = PV / RT
where:
P = pressure in atmospheres
V = volume in liters
T = temperature in kelvins
R = 0.08206 L atm/mol K (the ideal gas constant)
For NH3:
n(NH3) = (1.02 atm) (4.21 L) / (0.08206 L atm/mol K) (300 K)
n(NH3) = 0.177 mol
For HCl:
n(HCl) = (0.998 atm) (5.35 L) / (0.08206 L atm/mol K) (299 K)
n(HCl) = 0.204 mol
From the balanced equation, we can see that the stoichiometry of NH3 to HCl is 1:1, so NH3 is the limiting reactant since it has fewer moles. The balanced equation also tells us that the ratio of NH3 to NH4Cl is 1:1, so the number of moles of NH4Cl produced is also 0.177 mol.
To calculate the mass of NH4Cl produced, we need to use the molar mass of NH4Cl:
M(NH4Cl) = 14.01 g/mol (N) + 4(1.01 g/mol) (H) + 35.45 g/mol (Cl)
M(NH4Cl) = 53.49 g/mol
mass(NH4Cl) = n(NH4Cl) × M(NH4Cl)
mass(NH4Cl) = 0.177 mol × 53.49 g/mol
mass(NH4Cl) = 9.48 g
Therefore, the mass of NH4Cl produced is 9.48 g.
To determine the excess reactant, we can calculate how much of each reactant is consumed based on the stoichiometry of the reaction. Since NH3 and HCl react in a 1:1 ratio, we can assume that all of the NH3 is consumed, so we need to calculate the amount of HCl that is consumed:
n(HCl) consumed = n(NH3) = 0.177 mol
n(HCl) initially = (0.998 atm) (5.35 L) / (0.08206 L atm/mol K) (299 K)
n(HCl) initially = 0.204 mol
n(HCl) excess = n(HCl) initially - n(HCl) consumed
n(HCl) excess = 0.204 mol - 0.177 mol
n(HCl) excess = 0.027 mol
To convert this to a volume, we can use the ideal gas law:
V(HCl) excess = n(HCl) excess × RT / P
V(HCl) excess = 0.027 mol × (0.08206 L atm/mol K) (299 K) / (0.998 atm)
V(HCl) excess = 0.686 L
Therefore, the excess reactant is HCl, and the limiting reactant is NH3.