This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
You're on a roll — 26) Study the flow chart below and answer the questions that follow i) Identify gas X Gas X is Oxygen (O_2). ii) Write an equation for the reaction between ammonia and gas X 4NH_3(g) + 5O_2(g) Platinum catalyst 4NO(g) + 6H_2O(g) iii) Write an equation to show the formation of G and J 3NO_2(g) + H_2O(l) → 2HNO_3(aq) + NO(g) Here, G is HNO_3 (nitric acid) and J is NO (nitrogen (II) oxide). 27) All apparatus used during preparation nitric (V) acid are made of glass. Give a reason. Nitric (V) acid (HNO_3) is a highly corrosive and strong oxidizing acid. Glass is unreactive towards nitric acid, preventing corrosion of the apparatus and contamination of the product. 28) Sulphur (IV) oxide gas was bubbled through acidified potassium chromate (VI) solution and Iron (III) Sulphate solution. Explain the observations made in each case. i) With Potassium Chromate (VI) solution. Observation: The orange solution turns green*. Explanation: Sulphur (IV) oxide (SO_2) acts as a reducing agent. It reduces the orange chromate (VI) ions (CrO_4^2- or Cr_2O_7^2- in acidic solution) to green chromium (III) ions (Cr^3+). Cr_2O_7^2-(aq) + 3SO_2(g) + 2H^+(aq) → 2Cr^3+(aq) + 3SO_4^2-(aq) + H_2O(l) ii) Iron (III) Sulphate solution Observation: The yellow/brown solution turns pale green*. Explanation: Sulphur (IV) oxide (SO_2) acts as a reducing agent. It reduces the yellow/brown iron (III) ions (Fe^3+) to pale green iron (II) ions (Fe^2+). 2Fe^3+(aq) + SO_2(g) + 2H_2O(l) → 2Fe^2+(aq) + SO_4^2-(aq) + 4H^+(aq) Send me the next one 📸