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# AS chemistry practice question help- moles watch

1. 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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2. (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.
3. (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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