One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to NO: $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) form? \((\mathbf{c})\) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Using the molar mass of ammonia (\(NH_{3}\)) and oxygen (\(O_{2}\)), convert the given mass of each reactant to moles: Molar mass of \(NH_{3}\): N = \(14.01 \ g/mol\), H = \(1.01 \ g/mol\) Molar mass of \(NH_{3}\) = \(\displaystyle 1 \times 14.01 + 3 \times 1.01 = 17.04 \ g/mol\) Moles of ammonia \((NH_{3})\): $$ \text{moles of NH}_{3} = \frac{2.00\text{ g}}{17.04\text{ g/mol}}=0.1174 \text{ mol} $$ Molar mass of \(O_{2}\): O = \(16.00 \ g/mol\) Molar mass of \(O_{2}\) = \(2 \times 16.00 = 32.00 \ g/mol\) Moles of oxygen \((O_{2})\): $$ \text{moles of O}_{2}=\frac{2.50 \text{ g}}{32.00 \text{ g/mol}} =0.07813 \text{ mol} $$

Compare the mole-to-mole ratio of the reactants to the coefficients from the balanced chemical equation: $$ \frac{4 \mathrm{NH}_{3}(g)}{5 \mathrm{O}_{2}(g)} = \frac{0.1174 \text{ mol } NH_{3}}{0.07813 \text{ mol } O_{2}} = \frac{x}{0.07813 \text{ mol }} $$ Now, solve for x: $$ x = \frac{0.1174 \text{ mol } NH_{3}}{5} \times 0.07813 \text{ mol } O_{2} = 0.09686 \text{ mol } NH_{3} $$ Since the actual moles of \(NH_{3}\) (0.1174 mol) are greater than the required moles (0.09686 mol), oxygen (O₂) is the limiting reactant.

From the balanced chemical equation, we know the mole-to-mole ratios between the reactants and the products. Using the limiting reactant, we can find the moles of products formed: Moles of NO formed: $$ \frac{4 \mathrm{NO}(g)}{5 \mathrm{O}_{2}(g)} = \frac{x}{0.07813 \text{ mol } O_{2}} $$ Solve for x: $$ x = \frac{4}{5} \times 0.07813 \text{ mol } O_{2} = 0.06250 \text{ mol } \mathrm{NO} $$ Now find the mass of NO produced: Molar mass of NO: N = \(14.01 \ g/mol\), O = \(16.00 \ g/mol\) Molar mass of NO = \(14.01 + 16.00 = 30.01 \ g/mol\) Mass of NO formed: $$ \text{mass of NO} = 0.06250 \text{ mol } \times 30.01 \text{ g/mol} = 1.876 \text{ g} $$ Moles of H₂O formed: $$ \frac{6 \mathrm{H}_{2} \mathrm{O}(g)}{5 \mathrm{O}_{2}(g)} = \frac{y}{0.07813 \text{ mol } O_{2}} $$ Solve for y: $$ y = \frac{6}{5} \times 0.07813 \text{ mol } O_{2} = 0.09375 \text{ mol } \mathrm{H}_{2} \mathrm{O} $$ Now find the mass of H₂O produced: Molar mass of H₂O: H = \(1.01 \ g/mol\), O = \(16.00 \ g/mol\) Molar mass of H₂O = \(2 \times 1.01 + 16.00 = 18.02 \ g/mol\) Mass of H₂O formed: $$ \text{mass of H}_{2} \text{O} = 0.09375 \text{ mol } \times 18.02 \text{ g/mol} = 1.688 \text{ g} $$

Since oxygen is the limiting reactant, we will calculate how many moles of ammonia (NH₃) actually reacted: $$ \text{moles of NH}_{3} \text{ reacted} = 0.09686 \text{ mol} $$ Subtract this from the initial moles of ammonia: $$ \text{moles of NH}_{3} \text{ remaining} = 0.1174 \text{ mol } - 0.09686 \text{ mol } = 0.02054 \text{ mol} $$ Now, convert the moles of excess ammonia back to grams using its molar mass: $$ \text{mass of NH}_{3} \text{ remaining} = 0.02054 \text{ mol } \times 17.04 \text{ g/mol} = 0.3500 \text{ g} $$

To confirm that the Law of Conservation of Mass holds, the initial mass of reactants should equal the final mass of products and excess reactant: Initial mass of reactants: $$ 2.00 \text{ g } NH_{3} + 2.50 \text{ g } O_{2} = 4.50 \text{ g} $$ Final mass of products and excess reactant: $$ 1.876 \text{ g } \mathrm{NO} + 1.688 \text{ g } \mathrm{H}_{2} \mathrm{O} + 0.3500 \text{ g } NH_{3} = 4.914 \text{ g} $$ Since the initial mass of reactants (4.50 g) is equal to the final mass of products and excess reactant (4.914 g), the Law of Conservation of Mass holds true in our calculations.