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Chemistry - Gases (Thermal Decomposition of Metal Carbonates (A thermal…

- - - - In teh experiment below the initial volume of air was 200 cm3
      - The copper was heated and the air was repeatedly pushed from one gas syringe to the other. The copper started to turn black.
      - After a few minutes the heat was removed and the apparatus allowed to cool back down to room temperature.
      - The final volume was 162 cm3
    - - Explain why the copper started to turn black: The copper reacts with the oxygen in the air to turn copper oxide which is black.
      - Write an word equation for this reaction: copper + oxygen ---> carbon dioxide.
      - Calculate the volume of oxygen in the air: 162 cm3 + ... = 200 cm3, = 38cm3
    - - Describe how this apparatus can be used to determine the percentage of oxygen in the air: The rusting of iron requires both water and oxygen, so this reaction can be used in a similar way.
      - At the beginning of the experiment the length of the boiling tube that had air in it was 8.2 cm. At the end of the experiment the length of the boiling tube that had air in it was 6.1 cm. Calculate the percentage of oxygen in the sample of air.
      - 8.2 = 8 and 6.1 = 6
      - % of oxygen = 2/8 x 100
      - = 25%
    - - The apparatus initially contains 100 cm3 of air. The copper turnings are heated, while the air is pushed back and forth over them. The copper reacts with the oxygen in the air, to form copper oxide, and eventually all of the oxygen is used up. The apparatus is allowed to cool, and the final volume of gas which remains is recorded. Final volume = 79 cm3, therefore the air contained 21% oxygen.
    - - The rusting of iron requires both water and oxygen, so this reaction can be used in a similar way. The elastic band is used to mark the initial water level in the test tube. The apparatus is left for about a week. The damp iron wool will rust, and uses up all of the oxygen in the test tube as it does so. The new water level is marked with a pen. The volume of air inside the test tube can be calculated, before and after the rusting occurs. The volume should of decreased.
      - Initial reading - Today's reading = the amount of O2. The amount of oxygen divided by the initial reading times by 100.
  - - - The percentage of oxygen in the air can be determined by using phosphorus. The experiment below was set up:
      - At the beginning of the experiment the water level inside the bell jar was the same as the water level in the trough of water. The bell jar has a volume of 1000 cm3. During the experiment a bright light and white smoky solid were observed inside the bell jar. The white smoky solid is phosphorous oxide (P4 O10). When no further reaction was seen the water level had risen up inside the bell jar as shown in the diagram.
    - - Write a word equation: Phosphorus + oxygen ---> Phosphorus oxide
      - Explain why the water level inside the bell jar rose during the experiment: Because the phosphorus is burning therefore producing Co2. The more carbon dioxide produced the more water levels increase.
    - - Phosphorus is very reactive. It is stored under water, but if allowed to dry out it will spontaneously combust, reacting with oxygen, to form phosphorus oxide. The experiment can also be carried out using a small piece of phosphorous, instead pf the iron wool. Although this version is much quicker, it is also dangerous, so is not often used.
      - 4p + 5o2 ---> P4 O10