Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(\mathrm{~g})\), ammonia, \(\mathrm{NH}_{3}(\mathrm{~g})\), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), 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 \(20.0 \mathrm{~L} \mathrm{O}_{2}(\mathrm{~g})\) ? The volumes of all gases are measured at the same temperature and pressure.

Methane (CH4), ammonia (NH3), and oxygen (O2) react to form hydrogen cyanide (HCN) and gaseous water (H2O). The unbalanced chemical equation can be written as: CH4(g) + NH3(g) + O2(g) → HCN(g) + H2O(g) Now, balance the chemical equation: 2 CH4(g) + 2 NH3(g) + 3/2 O2(g) → 2 HCN(g) + 3 H2O(g) This can be written as fractions: CH4(g) + NH3(g) + 3/4 O2(g) → HCN(g) + 3/2 H2O(g) Or if you prefer whole numbers: 2 CH4(g) + 2 NH3(g) + 3 O2(g) → 4 HCN(g) + 6 H2O(g)

We are given 20.0 L of each reactant (CH4, NH3, and O2). To find the limiting reactant, we must find the mole ratio between the reactants and compare them with the balanced chemical equation. Since the volumes of all gases are measured at the same temperature and pressure, we can assume that their mole ratio remains the same, so no calculations regarding moles are necessary here. From the balanced chemical equation, the mole ratio between CH4, NH3, and O2 should be 2:2:3. Comparing the given volumes of each reactant: CH4:NH3:O2 20:20:20 Since the given ratios are not in the same proportion as the balanced chemical equation, there must be a limiting reactant. To find it, divide the volume of each reactant by the balanced equation's coefficient: For CH4: 20/2 = 10 For NH3: 20/2 = 10 For O2: 20/3 ~ 6.67 The smallest ratio is for O2, so O2 is the limiting reactant.

Now that we know O2 is the limiting reactant, we can calculate the amount of hydrogen cyanide produced. From the balanced chemical equation: 3 O2(g) -> 4 HCN(g), so the ratio between O2 and HCN is 3:4. Volume of HCN produced = (4/3) * Volume of O2 used Volume of HCN = (4/3) * 20.0 L Volume of HCN = 26.67 L So, the volume of hydrogen cyanide that can be obtained from the reaction of 20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g) is approximately 26.67 L.