Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(g)\) can be obtained from the reaction of $20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),$ and 20.0 \(\mathrm{L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

The given information indicates that \(\mathrm{CH}_{4}(g),\) \(\mathrm{NH}_{3}(g),\) and \(\mathrm{O}_{2}(g)\) react to produce hydrogen cyanide, \(\mathrm{HCN}(g),\) and gaseous water, \(\mathrm{H}_{2}\mathrm{O}(g).\) First, write the unbalanced equation and then balance it with the appropriate coefficients. Unbalanced equation: \[\mathrm{CH}_{4}(g) + \mathrm{NH}_{3}(g) + \mathrm{O}_{2}(g) \rightarrow \mathrm{HCN}(g) + \mathrm{H}_{2}\mathrm{O}(g)\] Now balance the equation: \[2\mathrm{CH}_{4}(g) + 6\mathrm{NH}_{3}(g) + 3\mathrm{O}_{2}(g) \rightarrow 6\mathrm{HCN}(g) + 6\mathrm{H}_{2}\mathrm{O}(g)\]

Since all of the volumes of gases are given and are measured at the same temperature and pressure, we can use the balanced chemical equation to determine the ratio between the reactants. The limiting reactant is the one that is used up first, as it determines the amount of product that can be formed. From the balanced equation, we have: \[2\mathrm{CH}_{4}(g) : 6\mathrm{NH}_{3}(g) : 3\mathrm{O}_{2}(g) \Rightarrow 1\mathrm{CH}_{4}(g) : 3\mathrm{NH}_{3}(g) : 1\frac{1}{2}\mathrm{O}_{2}(g)\] Comparing the molar ratios of the reactants, we have: \[\frac{20.0\ \mathrm{L\ CH}_{4}(g)}{1} : \frac{20.0\ \mathrm{L\ NH}_{3}(g)}{3} : \frac{20.0\ \mathrm{L\ O}_{2}(g)}{\frac{3}{2}} \Rightarrow 20.0 : 6.67 : 13.33\] Since the ratio of \(\mathrm{NH}_{3}\) is the smallest number, \(\mathrm{NH}_{3}\) is the limiting reactant.

As we found that the limiting reactant is \(\mathrm{NH}_{3},\) we will now use the stoichiometric coefficients to calculate the volume of \(\mathrm{HCN}(g)\) produced. From the balanced chemical equation, we can see that \(6\ \mathrm{moles}\) of \(\mathrm{NH}_{3}\) produce \(6\ \mathrm{moles}\) of \(\mathrm{HCN}.\) Therefore, the ratio is \(1 : 1,\) and the volume of \(\mathrm{HCN}\) produced will be equal to the volume of the limiting reactant, \(\mathrm{NH}_{3}\) used. Since we are given \(20.0\ \mathrm{L}\) of \(\mathrm{NH}_{3},\) the volume of \(\mathrm{HCN}\) produced is \(\frac{20.0\ \mathrm{L}}{3} = 6.67\ \mathrm{L}\) of \(\mathrm{HCN}(g)\). So, the volume of \(\mathrm{HCN}(g)\) obtained from the reaction of given volumes of \(\mathrm{CH}_{4}(g),\) \(\mathrm{NH}_{3}(g),\) and \(\mathrm{O}_{2}(g)\) is \(6.67\ \mathrm{L}\).