Chlorine gas, 0.600 mol, is passed into an excess of sodium hydroxide solution…
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Chlorine gas, , is passed into an excess of sodium hydroxide solution. In cold dilute alkali the products are sodium chloride and sodium hypochlorite. In hot concentrated alkali the products are sodium chloride and sodium chlorate. The equations are and . For the hot reaction find the amount of chloride, the amount of chlorate, the sodium hydroxide that is used and the mass of sodium chlorate. For the cold reaction find the mass of sodium hypochlorite. Take the molar masses as NaClO3 and NaOCl . Also say how many electrons are gained and lost per three Cl2 in the hot reaction.
Need a reminder? The key ideaShow
Redox: electrons lost equal electrons gained. In iodometry, 2 mol of thiosulphate react with 1 mol of iodine. In a disproportionation one species forms two products with higher and lower oxidation states. In a puzzle, follow the clues one at a time: gas tests, colours and the reagents of the standard conversions. At 273 K and 1 atm, 1 mol of gas fills 22.4 L.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Plan. Scale the equations to of chlorine.
Step 1. Hot reaction. The number of times the equation runs is . So chlorate is , chloride is , and sodium hydroxide is .
Step 2. Sodium chlorate. .
Step 3. Cold reaction. One mole of Cl2 gives one mole of NaOCl, so .
Step 4. Electrons. In 3 Cl2 (6 atoms), 5 atoms go from 0 to and gain 5 electrons. One atom goes from 0 to and loses 5 electrons.
Answer: of chloride, of chlorate, of base, of chlorate, of hypochlorite, and 5 electrons.
Check. Both reactions use of NaOH for each mole of Cl2: .
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