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tywower

  • one year ago

How many liters of oxygen gas can be produced if 28.7 grams of water decomposes at 294 Kelvin and 0.986 atmospheres? Show all of the work used to solve this problem. 2 H2O (l) yields 2 H2 (g) + O2 (g)

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  1. tywower
    • one year ago
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    @Abhisar

  2. tywower
    • one year ago
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    @sweetburger

  3. tywower
    • one year ago
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    @paki

  4. tywower
    • one year ago
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    @Kbug

  5. Abhisar
    • one year ago
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    Have you tried to do it on your own?

  6. tywower
    • one year ago
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    no, im having trouble remembering how to do this

  7. tywower
    • one year ago
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    @Abhisar

  8. Abhisar
    • one year ago
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    First you need to find the number of moles of oxygen formed. We can see that 1 mole of water forms 0.5 moles of oxygen so 28.7 grams of water i.e 28.7/18 =1.6 moles of water will form 0.8 moles of water. Any problem yet?

  9. Abhisar
    • one year ago
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    Now use the universal Gas law PV=nRT to find the volume of oxygen, Pressure is given 0.986 atm, T= 294 K V=nRT/P

  10. tywower
    • one year ago
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    no problem yet @Abhisar

  11. tywower
    • one year ago
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    1.01 times 2 plus 16.00 equals 18.02 grams per mole 28.7 grams of H2O times (1 mole divided by 18.02 grams) equals 1.59 moles of H2O water:oxygen = 2:1 is the mole ratio 1/2 times 1.59 equals 0.796 moles of O2 V = nRT/P P=0.986 atm V is not known n=0.797222 moles of O2 R=0.0821 T=294 K (0.986 atm)V=(0.797222 moles of O2)(0.0821)(294 K) V=19.5 Liters of O2

  12. tywower
    • one year ago
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    is this correct? @Abhisar

  13. Abhisar
    • one year ago
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    \(\huge \checkmark\)

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