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 of \(\mathrm{H}_{2} \mathrm{O}\) form? (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.

First, we need to calculate the number of moles of each reactant. To do this, we will divide the mass of the reactants by their respective molar masses. Molar masses: NH₃: 14.01 (N) + 3 * 1.01 (H) = 17.03 g/mol O₂: 2 * 16.00 (O) = 32.00 g/mol Number of moles of NH₃: \(= \frac {2.00 g}{17.03 g/mol} = 0.1174 \thinspace moles\) Number of moles of O₂: \(= \frac {2.50 g}{32.00 g/mol} = 0.07813 \thinspace moles\) Now that we have the moles of the reactants, we can determine the limiting reactant. According to the balanced chemical equation, 4 moles of NH₃ react with 5 moles of O₂. Therefore, the required mole ratio is: \(= \frac {4 \thinspace moles \thinspace NH_3}{5 \thinspace moles \thinspace O_2}\) Now, we compare the available moles of reactants to this ratio. Divide the available mole of each reactant by its corresponding mole value in the ratio: NH₃: \(= \frac {0.1174}{4} = 0.02935\) O₂: \(= \frac {0.07813}{5} = 0.01563\) Since O₂ has a smaller ratio, it is the limiting reactant.

We will use the number of moles of the limiting reactant (O₂) to find out the number of moles of the products formed (NO and H₂O). According to the balanced equation, 5 moles of O₂ react to form 4 moles of NO and 6 moles of H₂O. Number of moles of NO formed: \(= \frac {4 \thinspace moles \thinspace NO}{5 \thinspace moles \thinspace O_2} \times 0.07813 \thinspace moles \thinspace O_2 = 0.06250 \thinspace moles \thinspace NO\) Number of moles of H₂O formed: \(= \frac {6 \thinspace moles \thinspace H_2O}{5 \thinspace moles \thinspace O_2} \times 0.07813 \thinspace moles \thinspace O_2 = 0.09375 \thinspace moles \thinspace H_2O\) Now, multiply these values by their respective molar masses to find the mass of the formed products: Mass of NO formed: \(= 0.06250 \thinspace moles \thinspace NO \times \frac {30.01 \thinspace g}{1 \thinspace mol} = 1.875 \thinspace g \thinspace NO\) Mass of H₂O formed: \(= 0.09375 \thinspace moles \thinspace H_2O \times \frac {18.02 \thinspace g}{1 \thinspace mol} = 1.689 \thinspace g \thinspace H_2O\)

We know that O₂ is the limiting reactant, and thus, it will be completely consumed. We will now calculate how many moles of NH₃ were consumed during this reaction: Number of moles of NH₃ consumed: \(= \frac {4 \thinspace moles \thinspace NH_3}{5 \thinspace moles \thinspace O_2} \times 0.07813 \thinspace moles \thinspace O_2 = 0.06250 \thinspace moles \thinspace NH_3\) Now, we will find the number of moles of NH₃ remaining: Number of moles of NH₃ remaining: \(= 0.1174 \thinspace moles \thinspace NH_3 - 0.06250 \thinspace moles \thinspace NH_3 = 0.05490 \thinspace moles \thinspace NH_3\) Finally, we will calculate the mass of NH₃ remaining by multiplying it with the molar mass of NH₃: Mass of NH₃ remaining: \(= 0.05490 \thinspace moles \thinspace NH_3 \times \frac {17.03 \thinspace g}{1 \thinspace mol} = 0.935 \thinspace g \thinspace NH_3\)

According to the law of conservation of mass, the total mass of reactants should equal the total mass of products and unreacted excess reactant: Mass of reactants: \(= 2.00 \thinspace g \thinspace NH_3 + 2.50 \thinspace g \thinspace O_2 = 4.50 \thinspace g\) Add the masses of formed products NO and H₂O, and unreacted NH₃: Mass of products and remaining excess reactant: \(= 1.875 \thinspace g \thinspace NO + 1.689 \thinspace g \thinspace H_2O + 0.935 \thinspace g \thinspace NH_3 = 4.50 \thinspace g\) Since the total mass of reactants equals the total mass of products and the excess reactant, our calculations are consistent with the law of conservation of mass.