Final answer:
The theoretical yield of water when 40.0 g NH3 and 50.0 g O2 are mixed and allowed to react is calculated by first determining the limiting reactant, which is found to be oxygen (O2). Using stoichiometry and the balanced chemical equation, the theoretical yield of water is determined to be 1.87 moles of H2O.
Step-by-step explanation:
To calculate the theoretical yield of water from the reaction of ammonia with oxygen, we must first determine the limiting reactant, which will dictate the maximum amount of product that can be formed. Given the balanced chemical equation for the reaction:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
We will perform stoichiometric calculations to find the theoretical yield based on the initial amounts of reactants provided:
First, convert the mass of NH3 and O2 to moles using their respective molar masses (NH3 = 17.03 g/mol, O2 = 32.00 g/mol).
Calculate the moles of NH3:
40.0 g NH3 × (1 mol NH3 / 17.03 g NH3) = 2.35 moles NH3
Calculate the moles of O2:
50.0 g O2 × (1 mol O2 / 32.00 g O2) = 1.56 moles O2
Determine the limiting reactant using the stoichiometric coefficients from the balanced equation:
4 moles of NH3 react with 5 moles of O2, so the molar ratio is 4:5 (or NH3:O2).
Divide the moles of each reactant by its stoichiometric coefficient to find ratios:
NH3: 2.35 mol / 4 = 0.5875
O2: 1.56 mol / 5 = 0.3120
The lowest ratio corresponds to the limiting reactant, which is O2 in this case.
Use the molar ratio from the balanced equation to calculate moles of water produced by the limiting reactant (O2):
1.56 moles O2 × (6 moles H2O / 5 moles O2) = 1.87 moles H2O
The theoretical yield of water when 40.0 g NH3 and 50.0 g O2 react is 1.87 moles of H2O.