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    Hi,

    I've been doing some practice questions but there's no answers in the textbook. Please could someone go through how they'd approach these questions so that I can check if I'm on the right track.

    Thanks
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    (Original post by VioletPhillippo)
    Hi,

    I've been doing some practice questions but there's no answers in the textbook. Please could someone go through how they'd approach these questions so that I can check if I'm on the right track.

    Thanks
    First Screenshot:
    1a) P = 100 000 Pa and T = 298 K
    n = PV/RT
    n(total) = 174(.044)
    n (NO) = 69.6

    b) Moles of NH3=176.47
    Ratio between NH3 and NO2 is 1:1 therefore 176.47 x 46 = 8117.62
    Since there is only 80% yield: 8117.62 x 0.8= 6494 g
    Answer must be in kg: 6494/1000=6.5

    c) Moles of NO2: 0.543
    Ratio is 3:2, so moles of HNO3 is 0.543x2/3=0.362
    c=n/v, so 0.362/0.25dm3=1.448 mol dm-3

    d) NO2 contributes to acid rain / is an acid gas / forms HNO3 /NO2 is toxic / photochemical smog

    e) Ensure the ammonia is used up / ensure completereaction or combustionORMaximise the yield of nitric acid or products

    f) Neutralisation
    [this is from Jan 2013 AQA CHEM1 Q5]

    Second Screenshot:
    2b)Moles HCl = 0.12
    mol ZnCl2 = 0.12 / 2
    mass ZnCl2 = 0.06 × 136.4 = 8.18 g

    2c) Moles ZnCl2 = 10.7/136.4= 0.0784
    Since ratio of Zn: ZnCl2 is 1: 1 moles
    Zn = 0.0784
    Mass Zn reacting = 0.0784 × 65.4 = 5.13 g
    % purity of Zn = 5.13/5.68 × 100= 90.2%
    [This is from the AQA Jun 13 CHEM1 Mark scheme Q7]

    With regards to approaching these questions, it pretty much all moles calculations. You simply need to continue what you are currently doing; practice practice practice, it will all click...hopefully.
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    (Original post by john_jomcy98)
    First Screenshot:
    1a) P = 100 000 Pa and T = 298 K
    n = PV/RT
    n(total) = 174(.044)
    n (NO) = 69.6

    b) Moles of NH3=176.47
    Ratio between NH3 and NO2 is 1:1 therefore 176.47 x 46 = 8117.62
    Since there is only 80% yield: 8117.62 x 0.8= 6494 g
    Answer must be in kg: 6494/1000=6.5

    c) Moles of NO2: 0.543
    Ratio is 3:2, so moles of HNO3 is 0.543x2/3=0.362
    c=n/v, so 0.362/0.25dm3=1.448 mol dm-3

    d) NO2 contributes to acid rain / is an acid gas / forms HNO3 /NO2 is toxic / photochemical smog

    e) Ensure the ammonia is used up / ensure completereaction or combustionORMaximise the yield of nitric acid or products

    f) Neutralisation
    [this is from Jan 2013 AQA CHEM1 Q5]

    Second Screenshot:
    2b)Moles HCl = 0.12
    mol ZnCl2 = 0.12 / 2
    mass ZnCl2 = 0.06 × 136.4 = 8.18 g

    2c) Moles ZnCl2 = 10.7/136.4= 0.0784
    Since ratio of Zn: ZnCl2 is 1: 1 moles
    Zn = 0.0784
    Mass Zn reacting = 0.0784 × 65.4 = 5.13 g
    % purity of Zn = 5.13/5.68 × 100= 90.2%
    [This is from the AQA Jun 13 CHEM1 Mark scheme Q7]

    With regards to approaching these questions, it pretty much all moles calculations. You simply need to continue what you are currently doing; practice practice practice, it will all click...hopefully.
    Thank you so much, I really appreciate it. I've had a look through and I have got most of them right but I just wanted to double check!
 
 
 
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