Part C Determine the limiting reactant. \[ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \] Express your answer as a chemical formula. \( \mathrm{O}_{2} \) Submit Previous Answers Correct Correct Part D Calculate the grams of product in parentheses that would be produced. \( \left(\mathrm{H}_{2} \mathrm{O}\right) \) Express your answer with the appropriate units. \[ \mathrm{m}\left(\mathrm{H}_{2} \mathrm{O}\right)=10.14 \mathrm{~g} \] Submit Previous Answers Request Answer Incorrect; Try Again; 5 attempts remaining

Given that we already determined that the limiting reactant is \[ \mathrm{O}_{2}, \] we work only with the amount of \(\mathrm{O}_{2}\) available. In this reaction \[ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \] the stoichiometric ratio shows that 5 moles of \(\mathrm{O}_{2}\) produce 6 moles of \(\mathrm{H}_{2}\mathrm{O}\). That is, for every mole of \(\mathrm{O}_{2}\) that reacts, the number of moles of water produced is \[ n(\mathrm{H}_{2}\mathrm{O}) = \left(\frac{6}{5}\right) n(\mathrm{O}_{2}). \] Assume that the mass of \(\mathrm{O}_{2}\) provided is \(15.00\) g (this is the value that leads to the numbers in our solution). First, calculate the moles of \(\mathrm{O}_{2}\): \[ n(\mathrm{O}_{2}) = \frac{m(\mathrm{O}_{2})}{M(\mathrm{O}_{2})} = \frac{15.00\ \mathrm{g}}{32.00\ \mathrm{g/mol}} = 0.46875\ \mathrm{mol}. \] Next, we determine the moles of water formed using the stoichiometric ratio: \[ n(\mathrm{H}_{2}\mathrm{O}) = \left(\frac{6}{5}\right) (0.46875\ \mathrm{mol}) = 0.5625\ \mathrm{mol}. \] Now, we convert the moles of water to grams using the molar mass of water. (We will use \(M(\mathrm{H}_{2}\mathrm{O}) \approx 18.02\ \mathrm{g/mol}\); note that slight variations in this constant may change the final answer in the last decimal place.) \[ m(\mathrm{H}_{2}\mathrm{O}) = n(\mathrm{H}_{2}\mathrm{O}) \times M(\mathrm{H}_{2}\mathrm{O}) = 0.5625\ \mathrm{mol} \times 18.02\ \mathrm{g/mol} \approx 10.13\ \mathrm{g}. \] Thus, the mass of water produced is \[ 10.13\, \mathrm{g}. \]