For the reaction \(\mathrm{NH}_{3}+\mathrm{NO} \rightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}:\) (a) Balance the equation. (b) If you react \(15.0 \mathrm{~g} \mathrm{NH}_{3}\) with \(22.0 \mathrm{~g} \mathrm{NO}\) and you produce \(13.3 \mathrm{~g} \mathrm{~N}_{2}\), what is your percent yield?

To balance the equation, we need to adjust the coefficients of the reactants and products, so that the number of atoms of each element is equal on both sides of the reaction. The given reaction is: \(\mathrm{NH}_{3} + \mathrm{NO} \rightarrow \mathrm{N}_{2} + \mathrm{H}_{2}\mathrm{O}\) There are two nitrogen atoms on the right side while there is only one nitrogen atom on each of the reactants. Changing the coefficient of \(\mathrm{NH}_{3}\) to 2 will balance the nitrogen atoms: \[\underline{\bold{2}}\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+\mathrm{H}_{2}\mathrm{O}\] Now, amend the coefficients of water to balance the hydrogen atoms: \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+\underline{\bold{3}}\mathrm{H}_{2}\mathrm{O}\) The balanced equation is: \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+3\mathrm{H}_{2}\mathrm{O}\)

Given amounts of reactants: - 15.0 g of \(\mathrm{NH}_{3}\) - 22.0 g of \(\mathrm{NO}\) First, calculate the moles of each reactant. Molar mass of \(\mathrm{NH}_{3} = 14.01 (\mathrm{N}) + 3 \times 1.01(\mathrm{H}) = 17.03 \text{ g/mol}\) Moles of \(\mathrm{NH}_{3} = \frac{15.0 \text{ g}}{17.03 \text{ g/mol}} = 0.881 \text{ mol}\) Molar mass of \(\mathrm{NO} = 14.01 (\mathrm{N}) + 16.00 (\mathrm{O}) = 30.01 \text{ g/mol}\) Moles of \(\mathrm{NO} = \frac{22.0 \text{ g}}{30.01 \text{ g/mol}} = 0.733 \text{ mol}\) Now, divide the moles of each reactant by their respective coefficients in the balanced equation and find the lowest ratio: \(\frac{\text{moles of NH}_{3}}{2} = \frac{0.881}{2} = 0.441\) \(\frac{\text{moles of NO}}{1} = \frac{0.733}{1} = 0.733\) As 0.441 is the lowest ratio, \(\mathrm{NH}_{3}\) is the limiting reactant.

Using the stoichiometry of the balanced equation, calculate the theoretical amount of \(\mathrm{N}_{2}\) that can be produced: \(2\mathrm{NH}_{3}+\mathrm{NO}\rightarrow \mathrm{N}_{2}+3\mathrm{H}_{2}\mathrm{O}\) From the balanced equation, two moles of \(\mathrm{NH}_{3}\) give one mole of \(\mathrm{N}_{2}\). So, we can calculate the theoretical moles of \(\mathrm{N}_{2}\): Theoretical moles of \(\mathrm{N}_{2} = \frac{1}{2} \times \text{moles of NH}_{3}\) Theoretical moles of \(\mathrm{N}_{2} = \frac{1}{2} \times 0.881 = 0.441 \text { mol}\) Now, convert moles of \(\mathrm{N}_{2}\) to mass using its molar mass: Molar mass of \(\mathrm{N}_{2} = 2 \times 14.01 = 28.02 \text{ g/mol}\) Theoretical yield of \(\mathrm{N}_{2} = 0.441 \text{ mol} \times 28.02 \text{ g/mol} = 12.35 \text{ g}\)

We are given that the actual yield of \(\mathrm{N}_{2}\) is 13.3 g. To calculate the percent yield, use the formula: Percent yield = \(\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\) Percent yield = \(\frac{13.3 \text{ g}}{12.35 \text{ g}} \times 100\% = 107.7\%\) This value is greater than 100%, which means there is likely an error in the actual yield provided. For the given amounts of reactants, the obtained yield of \(\mathrm{N}_{2}\) should not exceed the theoretical yield. Nonetheless, the calculated percent yield based on the given data is 107.7%.