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A2 Chemistry: Calculating concentration given Kc question help watch

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    I'm having a terrible time with this question. The answer is 0.897 mol dm–3 but I'm stuck at getting 0.93 for some reason. Our teacher told us that she is giving the same type of question in our test tomorrow so I am absolutely terrified. Here it is:

    Phosphorus pentachloride decomposes as shown:
    PCl5 (g) -----> PCl3 (g) + Cl2 (g)
    Given that the equilibrium constant for this reaction is 11.5 mol dm-3 at 300 K, what is the equilibrium concentration of PCl3 when the concentration of PCl5 at equilibrium is 0.07 mol dm-3

    Ideas? Any help would be appreciated
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    11.5=[Cl2] * [PCl3]/0.07

    let [PCl3] = x

    11.5* 0.07 = x^2

    x = sqrt (11.5* 0.07)

    x = 0.897 mol/dm3
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    (Original post by Maker)
    11.5=[Cl2] * [PCl3]/0.07

    let [PCl3] = x

    11.5* 0.07 = x^2

    x = sqrt (11.5* 0.07)

    x = 0.897 mol/dm3
    why did you square the x? and what happened to the concentration of Cl2? why did it disappear from the expression?
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    (Original post by Tara1208)
    why did you square the x? and what happened to the concentration of Cl2? why did it disappear from the expression?
    Because the products are in equal concentrations

    x*x = X^2
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    (Original post by Lord Nutter)
    Because the products are in equal concentrations

    x*x = X^2
    How did you know that?
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    (Original post by Tara1208)
    How did you know that?
    Look at the equation for the decomposition ...
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    (Original post by charco)
    Look at the equation for the decomposition ...
    i still don't get it.. 😣 i get that the number of moles of PCL3 and Cl2 is same..but still how come their concentration would be same at equilibrium? 😔
    please can you elaborate
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    (Original post by Tara1208)
    i still don't get it.. 😣 i get that the number of moles of PCL3 and Cl2 is same..but still how come their concentration would be same at equilibrium? 😔
    please can you elaborate
    PCl5 --> PCl3 + Cl2
    1mol --------------> x mol + x mol
 
 
 
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