Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If \(5.00 \times 10^{3} \mathrm{kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2},\) and \(\mathrm{CH}_{4}\) are reacted, what mass of HCN and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Using the molar mass, we will calculate the number of moles of each reactant available: NH3: \(1 NH_3 = 14.01 (N) + 3 \cdot 1.01 (H) = 17.03 \, g/mol\) O2: \(1 O_2 = 2 \cdot 16.00 (O) = 32.00 \, g/mol\) CH4: \(1 CH_4 = 12.01 (C) + 4 \cdot 1.01 (H) = 16.04 \, g/mol \) Now, we will convert the given mass of each reactant to moles: NH3: \(\frac{5.00 \times 10^3 \, kg}{17.03 \, g/mol} = 2.94 \times 10^5 \, moles\) O2: \(\frac{5.00 \times 10^3 \, kg}{32.00 \, g/mol} = 1.56 \times 10^5 \, moles\) CH4: \(\frac{5.00 \times 10^3 \, kg}{16.04 \, g/mol} = 3.12 \times 10^5 \, moles\)

Now we need to determine the limiting reactant, the reactant that will be completely used up in the reaction, by comparing the moles of each reactant with the stoichiometry given by the balanced chemical equation. We do this by finding the reactant that has the lowest ratio of moles available to the stoichiometric coefficients: NH3: \(\frac{2.94 \times 10^5}{2} = 1.47 \times 10^5\) O2: \(\frac{1.56 \times 10^5}{3} = 5.20 \times 10^4\) CH4: \(\frac{3.12 \times 10^5}{2} = 1.56 \times 10^5\) From the calculations above, O2 has the lowest ratio, which means it is the limiting reactant.

With O2 as the limiting reactant, we can now use stoichiometry to determine the moles of HCN and H2O produced: Moles of HCN produced: \(1.56 \times 10^5 \, moles \, O_2 \cdot \frac{2 \, moles \, HCN}{3 \, moles \, O_2} = 1.04 \times 10^5 \, moles\) Moles of H2O produced: \(1.56 \times 10^5 \, moles \, O_2 \cdot \frac{6 \, moles \, H_2O}{3 \, moles \, O_2} = 3.12 \times 10^5 \, moles\) Now, let's calculate the mass of each product using their molar masses: HCN: \(1.04 \times 10^5 \, moles \cdot 27.03 \, g/mol = 2.81 \times 10^6 \, g = 2.81 \times 10^3 \, kg\) (The molar mass of HCN is calculated as \(\, 1C + 1N + 1H = 12.01 + 14.01 + 1.01 = 27.03 \, g/mol \) ) H2O: \(3.12 \times 10^5 \, moles \cdot 18.02 \, g/mol = 5.62 \times 10^6 \, g = 5.62 \times 10^3 \, kg\) (The molar mass of H2O is calculated as \(\, 2H + 1O = 2 \cdot 1.01 + 16.00 = 18.02 \, g/mol \) ) Therefore, based on a 100% yield, 2.81 x 10^3 kg of HCN and 5.62 x 10^3 kg of H2O will be produced from the given amounts of reactants.