Final answer:
To determine the limiting reagent, calculate the moles of each reactant and compare it to the stoichiometry. NH3 is the limiting reagent and O2 is in excess. No unreacted NH3 is left over. The volume of NO produced at NTP is 22.4 L. The mass of water produced is 27.03 g.
Step-by-step explanation:
To determine the limiting reagent, we need to calculate the moles of each reactant and compare it to the stoichiometry of the balanced equation. The balanced equation for the reaction is:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
To calculate the moles of each reactant, divide the given mass by the molar mass. The molar mass of ammonia (NH3) is 17.03 g/mol and the molar mass of oxygen (O2) is 32.00 g/mol.
Moles of NH3 = 17 g / 17.03 g/mol = 1 mol
Moles of O2 = 45 g / 32.00 g/mol = 1.41 mol
From the balanced equation, we can see that 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O. So, the ratio of moles of NH3 to O2 is 4:5.
In this case, the ratio of moles of NH3 to O2 in the reaction is 1:1.41, which indicates that NH3 is the limiting reagent and O2 is in excess.
To calculate the moles of unreacted reactant left over, we need to compare the moles of the limiting reagent (NH3) with the stoichiometry of the balanced equation.
Moles of NO produced = 1 mol of NH3 x (4 mol NO / 4 mol NH3) = 1 mol
Since the stoichiometry of the balanced equation is 1:1, the number of moles of unreacted NH3 left over is 0 mol.
To calculate the volume of NO produced at NTP (standard temperature and pressure), we need to use the ideal gas law equation PV = nRT, where P = pressure, V = volume, n = moles, R = ideal gas constant, and T = temperature in Kelvin.
At NTP, the pressure is 1 atmosphere and the temperature is 273 Kelvin. The ideal gas constant (R) is 0.0821 L·atm/(mol·K).
Volume of NO = moles of NO x (R x T / P) = 1 mol x (0.0821 L·atm/(mol·K) x 273 K / 1 atm) = 22.4 L
To calculate the mass of water produced, we need to use the stoichiometry of the balanced equation.
Moles of H2O produced = 1 mol of NH3 x (6 mol H2O / 4 mol NH3) = 1.5 mol
The molar mass of water (H2O) is 18.02 g/mol.
Mass of H2O produced = moles of H2O x molar mass = 1.5 mol x 18.02 g/mol = 27.03 g