The balanced chemical equation for the reaction between nitrogen trihydride (NH3) and diatomic oxygen (O2) to produce water (H2O) and nitrogen monoxide (NO) is:
4 NH3 + 3 O2 → 2 N2 + 6 H2O
We can use the given masses of NH3 and O2 to determine which is the limiting reactant, and then use the stoichiometry of the balanced equation to find the number of molecules of water produced.
First, we need to convert the masses of NH3 and O2 to moles using their respective molar masses:
NH3: 17.88 g / 17.03 g/mol = 1.050 mol
O2: 11.9 g / 32.00 g/mol = 0.372 mol
Next, we can use the stoichiometry of the balanced equation to determine which reactant is limiting. The balanced equation tells us that 4 moles of NH3 react with 3 moles of O2 to produce 6 moles of H2O. Therefore, the ratio of NH3 to O2 needed for complete reaction is:
4 mol NH3 / 3 mol O2 = 1.33 mol NH3 / mol O2
Since the actual ratio of NH3 to O2 is:
1.050 mol NH3 / 0.372 mol O2 = 2.82 mol NH3 / mol O2
We can see that NH3 is present in excess, and O2 is the limiting reactant.
Using the mole ratio from the balanced equation, we can now calculate the number of moles of H2O produced from the given amount of O2:
0.372 mol O2 × (6 mol H2O / 3 mol O2) = 0.744 mol H2O
Finally, we can use Avogadro's number to convert the number of moles of H2O to molecules:
0.744 mol H2O × (6.022 × 10^23 molecules/mol) = 4.47 × 10^23 molecules of H2O
Therefore, the number of molecules of water that may be produced is approximately 4.47 × 10^23.