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.
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Here are the solutions for part (e): e) A student is given the following mixtures: e) (i) What do you observe in the tube containing K_2Cr_2O_7 solution? When sodium sulfite (Na_2SO_3) reacts with hydrochloric acid (HCl), sulfur dioxide (SO_2) gas is produced. Sulfur dioxide is a reducing agent. Potassium dichromate (K_2Cr_2O_7) solution is an orange oxidizing agent. When SO_2 gas is passed through the K_2Cr_2O_7 solution, the dichromate ions (Cr_2O_7^2-) are reduced to green chromium(III) ions (Cr^3+). The observation is that the orange solution turns green. e) (ii) Identify the gas produced in the first test tube A. The reaction between sodium sulfite (Na_2SO_3) and hydrochloric acid (HCl) produces sulfur dioxide gas. Na_2SO_3(s) + 2HCl(aq) → 2NaCl(aq) + H_2O(l) + SO_2(g) The gas produced is sulfur dioxide (SO_2. e) (iii) State on physical test for the gas produced in test tube A. Sulfur dioxide is a colorless gas with a characteristic pungent smell. A physical test for sulfur dioxide is its pungent/choking/suffocating smell. That's 2 down. 3 left today — send the next one.