Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g)\), ammonia, \(\mathrm{NH}_{3}\left(\mathrm{~g}\right.\) ), 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 \(\mathrm{HCN}(\mathrm{g})\) 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.

To write the balanced chemical equation for the reaction, we need to identify the products and balance the number of atoms on both sides of the equation: From the exercise, we know that hydrogen cyanide (HCN), methane (CH4), ammonia (NH3), oxygen (O2), and gaseous water (H2O) are involved in the reaction. \[ CH_4(g) + NH_3(g) + O_2(g) \rightarrow HCN(g) + H_2O(g) \] Now, we need to balance the equation: \[ 2 CH_4(g) + 2 NH_3(g) + 3 O_2(g) \rightarrow 2 HCN(g) + 6 H_2O(g) \] The balanced chemical equation for the reaction is: \[ 2 CH_4(g) + 2 NH_3(g) + 3 O_2(g) \rightarrow 2 HCN(g) + 6 H_2O(g) \]

To find the volume of hydrogen cyanide produced, we will use the stoichiometric coefficients from the balanced chemical equation. Given volumes of reactants: 1. 20.0 L CH4 2. 20.0 L NH3 3. 20.0 L O2 At the same temperature and pressure, the ratio of the number of moles of gases is the same as the ratio of their volumes: Let's find the limiting reactant by comparing the volume ratios of reactants with the stoichiometric coefficients: 1. For CH4: \(\frac{20.0}{2} = 10 \) 2. For NH3: \(\frac{20.0}{2} = 10 \) 3. For O2: \(\frac{20.0}{3} = 6.67 \) It appears that O2 (oxygen) is the limiting reactant in this reaction. Now, we can find the volume of hydrogen cyanide (HCN) formed by using the stoichiometric coefficient of O2: The stoichiometric coefficient ratio between HCN and O2 is \(\frac{2}{3}\). Therefore, volume of HCN produced = \( 20.0 L\ O_2 \ \times \frac{2}{3} = 13.33 L\)

The volume of HCN(g) 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 13.33 L.