One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO} :\) $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ 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? (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.

We need to determine the number of moles of NH3 and O2 from their given mass. Use the formula: moles = mass (in grams) / molar mass (in g/mol). Molar mass of \(\mathrm{NH}_{3}\) is 17.03 g/mol, and for \(\mathrm{O}_{2}\), it is 32.00 g/mol. Number of moles of \(\mathrm{NH}_{3} =\frac{2.00\,\mathrm{g}}{17.03\,\mathrm{g/mol}} =0.117\,\mathrm{mol}\) Number of moles of \(\mathrm{O}_{2} =\frac{2.50\,\mathrm{g}}{32.00\,\mathrm{g/mol}} =0.0781\,\mathrm{mol}\)

To find the limiting reactant, we need to examine the stoichiometric ratios in the balanced equation. We will compare the ratios of the initial moles to the coefficients in the balanced equation. Ratios: \(\frac{\mathrm{mol\,NH}_{3}}{4} = \frac{0.117}{4} = 0.0292\) \(\frac{\mathrm{mol\,O}_{2}}{5} = \frac{0.0781}{5} = 0.0156\) Since the ratio of oxygen is smaller than that of ammonia, \(\mathrm{O}_{2}\) is the limiting reactant.

Now that we know the limiting reactant, we can calculate the mass of the products \(\mathrm{NO}\) and \(\mathrm{H}_{2}\mathrm{O}\). Use stoichiometric relationships (coefficients in the balanced equation) to find the moles of products formed by the limiting reactant. Moles of \(\mathrm{NO}\) formed: \(\mathrm{mol\, O}_{2} \times(\frac{4\,\text{mol}\,\mathrm{NO}}{5\,\text{mol}\,\mathrm{O}_{2}})= 0.0781 \times(\frac{4}{5})=0.0625\,\text{mol}\,\mathrm{NO}\) Moles of \(\mathrm{H}_{2}\mathrm{O}\) formed: \(\mathrm{mol\, O}_{2} \times(\frac{6\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}}{5\,\text{mol}\,\mathrm{O}_{2}})= 0.0781 \times(\frac{6}{5})=0.0936\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}\) We can now convert the moles of products to grams by multiplying with the molar mass of the respective product: Mass of \(\mathrm{NO} = 0.0625\,\mathrm{mol} \times 30.01\,\mathrm{g/mol} = 1.88\,\mathrm{g}\) Mass of \(\mathrm{H}_{2}\mathrm{O} = 0.0936\,\mathrm{mol} \times 18.02\,\mathrm{g/mol} = 1.69\,\mathrm{g}\)

Now that we found our limiting reactant and the products formed, we can compute the mass of the excess reactant, \(\mathrm{NH}_{3}\), that will remain after the limiting reactant is completely consumed. Moles of \(\mathrm{NH}_{3}\) consumed: \(\mathrm{mol\, O}_{2} \times(\frac{4\,\text{mol}\,\mathrm{NH}_{3}}{5\,\text{mol}\,\mathrm{O}_{2}})= 0.0781 \times(\frac{4}{5})=0.0625\,\text{mol}\,\mathrm{NH}_{3}\) Moles of \(\mathrm{NH}_{3}\) remaining: \(0.117 \, \mathrm{mol} - 0.0625 \, \mathrm{mol} = 0.0545 \, \mathrm{mol}\) Mass of \(\mathrm{NH}_{3}\) remaining: \(0.0545\,\mathrm{mol} \times 17.03\,\mathrm{g/mol} = 0.928\, \mathrm{g}\)

To check if our calculations are consistent with the law of conservation of mass, we will ensure that the mass of the reactants equals the mass of the products and the excess reactant remaining. Mass of reactants: \(2.00\,\mathrm{g\,of\, NH}_{3}+2.50\,\mathrm{g\,of\, O}_{2}=4.50\,\mathrm{g}\) Mass of products and excess reactant: \(1.88\,\mathrm{g\,of\, NO} + 1.69\,\mathrm{g\,of\, H}_{2}\mathrm{O} + 0.928\,\mathrm{g\,of\, NH}_{3} \approx 4.50\,\mathrm{g}\) The mass of reactants and products are approximately equal, which is consistent with the law of conservation of mass.