yeah this is AQA GCSE chemistry(Original post by shady2.0)
This topic is for GSCE right?
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TheConfusedMedic
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 16022016 19:02

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 16022016 19:04
(Original post by vickie89uk)
Ah ok I think I understand thank you.... That makes sense..... Thank you so much you have the patience of a saint...
Sure there's many tsr CHEM students laughing at my idiocy 
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 16022016 19:10
(Original post by shady2.0)
This topic is for GSCE right? 
vickie89uk
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 16022016 19:12
(Original post by samb1234)
No worries, best of luck with your gcses 
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 16022016 19:16
(Original post by vickie89uk)
Thank you so much with a bit of look I won't be spending an hour on a single question..... As long as I can get a B I'll be ok I'm going to keep practising this.... What I failed to realise was I needed to make the atoms on the left balanced with the right and then work out the masses and then the products we started with and produced... Only took all day thank you so much for teaching me and not just giving me the answer 
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 16022016 20:04
(Original post by samb1234)
No worries, i find it's much more useful to lead people through as then they tend to be able to reproduce it in the exam better if they actually understand what they're doing, best of luck 
Chemist1997
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 16022016 20:05
The reaction of hexaamine cobalt (II) chloride (CoCl2•6NH3) with hydrogen peroxide to form cobalt (III) hydroxide pentaamine chloride (CoCl2(OH)•5NH3). Cana anyone work out the equation?

vickie89uk
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 16022016 21:23
Attachment 504283504285H
(Original post by samb1234)
If you have Na2co3 and h2o there must have been 2 nahco3 in the first place, make sense?
Then we got 16g for Mgo I divided that by mad 16/36
= 0.4444'
Co2 is 44g I don't know where I go from here but I included my workings⬇️
44/0.4444 would work but it doesn't
The answer is 17.6g 
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 16022016 21:29
(Original post by vickie89uk)
Attachment 504283504285H
So I've got this question ⬆️I've balanced the equation (it was balanced already) I've decided what we started with by the mass 33.6g/84 = 0.4 for mgco3
Then we got 16g for Mgo I divided that by mad 16/36
= 0.4444'
Co2 is 44g I don't know where I go from here but I included my workings⬇️
44/0.4444 would work but it doesn't
The answer is 17.6g
0.4 (from the left) X 44g from co2 = 17.6g I hope that's right
(Original post by samb1234)
No worries, i find it's much more useful to lead people through as then they tend to be able to reproduce it in the exam better if they actually understand what they're doing, best of luck 
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 16022016 21:45
(Original post by vickie89uk)
I got it haha
0.4 (from the left) X 44g from co2 = 17.6g I hope that's right 
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 16022016 21:53
(Original post by samb1234)
Sounds good to me
Yay I so happy I finally get it..... I was looking at the page and it suddenly clicked as I thought why are you multiply Mgo and carbon when their both on the right side.... 
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 16022016 22:00
There is one final complication. If you had more than 1 reactant, you would need to work out what the limiting reagent is
(Original post by vickie89uk)
Because I have to multiply the 0.4 from the left by the mass of carbon because carbon came from the left... The mass I get for Mgo is irrelevant to the final sum
Yay I so happy I finally get it..... I was looking at the page and it suddenly clicked as I thought why are you multiply Mgo and carbon when their both on the right side.... 
vickie89uk
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 16022016 22:08
(Original post by samb1234)
There is one final complication. If you had more than 1 reactant, you would need to work out what the limiting reagent is 
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 17022016 14:06
(Original post by vickie89uk)
Oh my god how would you even do that. I don't even know what that means... Haha but I'm getting it learnt more in 6 hours here than I have in half a year in class
Question: Sodium metal reacts with chlorine gas to form a white crystalline soild of sodium chloride. John reacts 24.2g of sodium metal with 1 mole of chlorine gas, and he produces 26.4g of sodium chloride. Calculate his % yield.
Step 1: Write out the full balanced symbol equation, so firstly you need to work out the formula of the product. Sodium is a group one metal so will form a 1+ ion, and chlorine is in group 7 so will form a 1 ion. Therefore the formula of the product is NaCl. We know that chlorine is diatomic so the initial, unbalanced equation is:
Na +Cl2 > NaCl
To balance we need there to be the same amount of each element on both sides of the equation . At the moment there are 2 chlorines on the left and only one on the right, so we need 2 NaCl:
Na +Cl2 > 2NaCl
We now have the right number of chlorines on both sides, but we now have 2 sodiums on the right and only one on the left, so we need 2 Na so the final, balanced equation would be:
2Na +Cl2 > 2NaCl
Step 2
Now that we have the equation, we first want to work out the number of moles we have of every reagent. In this example, we are told that we have one mole of chlorine. We have 24.2g of sodium metal, and one mole of sodium would weigh 23g. Therefore we have:
24.2/23 = 1.05... moles of sodium, and 1 mole of chlorine (given in question in this example, but if it wasn't you just work out the number of moles of all the reagents makes no difference).
Step 3
We now need to look at the balanced equation and the number of moles we have to work out the limiting reagent. What does this mean? Well imagine you were making a cake, and each cake took 3eggs and a kg of flour. If you only had 3kg of flour, even if you had 100 eggs you would still only be able to make 3 cakes, so the flour would be the limiting reagent.
So we know from the balanced equation that for every 2 moles of sodium reacting we need 1 mole of chlorine. However, you should clearly be able to see that we do not have twice as much sodium as there is chlorine. Even though we have 1 mole of chlorine, since we do not have 2 moles of sodium the chlorine is in excess.
Step 4
Now that we have identified the sodium as the limiting reagent, we are able to establish the expected yield. Looking back at the original equation, the number of moles of Na is the same as the number of moles of NaCl. Therefore if we have 1.05... moles of Na, we would theoretically expect to get 1.05... moles of NaCl.
Therefore the expected yield in grams would be (mass of one mole) x (number of moles we have). The expected yield is therefore 1.05... x (23+35.5) = 61.6g
Step 5
Now we have the expected yield we can work out the % yield.
% yield = mass we got/ expected mass x 100
= 26.4/61.6 *100 = 42.9%
Hope that helped 
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 17022016 14:08
(Original post by surina16)
yeah this is AQA GCSE chemistry 
TheConfusedMedic
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 17022016 14:35
(Original post by samb1234)
I don't know if you now fully get it but you might find my post helpful idk
Your explanations are amazing! 
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 17022016 14:53
(Original post by surina16)
Solved your question without looking and then checked it with the post just to make sure
Your explanations are amazing! 
vickie89uk
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 17022016 15:21
(Original post by samb1234)
Right I'm going to go for a bit of a different approach, I'm going to do a fully worked example explaining each stage as I go to hopefully allow you to answer anything they ask. Note that I'm just making this question up it doesn't come from anywhere.
Question: Sodium metal reacts with chlorine gas to form a white crystalline soild of sodium chloride. John reacts 24.2g of sodium metal with 1 mole of chlorine gas, and he produces 26.4g of sodium chloride. Calculate his % yield.
Step 1: Write out the full balanced symbol equation, so firstly you need to work out the formula of the product. Sodium is a group one metal so will form a 1+ ion, and chlorine is in group 7 so will form a 1 ion. Therefore the formula of the product is NaCl. We know that chlorine is diatomic so the initial, unbalanced equation is:
Na +Cl2 > NaCl
To balance we need there to be the same amount of each element on both sides of the equation . At the moment there are 2 chlorines on the left and only one on the right, so we need 2 NaCl:
Na +Cl2 > 2NaCl
We now have the right number of chlorines on both sides, but we now have 2 sodiums on the right and only one on the left, so we need 2 Na so the final, balanced equation would be:
2Na +Cl2 > 2NaCl
Step 2
Now that we have the equation, we first want to work out the number of moles we have of every reagent. In this example, we are told that we have one mole of chlorine. We have 24.2g of sodium metal, and one mole of sodium would weigh 23g. Therefore we have:
24.2/23 = 1.05... moles of sodium, and 1 mole of chlorine (given in question in this example, but if it wasn't you just work out the number of moles of all the reagents makes no difference).
Step 3
We now need to look at the balanced equation and the number of moles we have to work out the limiting reagent. What does this mean? Well imagine you were making a cake, and each cake took 3eggs and a kg of flour. If you only had 3kg of flour, even if you had 100 eggs you would still only be able to make 3 cakes, so the flour would be the limiting reagent.
So we know from the balanced equation that for every 2 moles of sodium reacting we need 1 mole of chlorine. However, you should clearly be able to see that we do not have twice as much sodium as there is chlorine. Even though we have 1 mole of chlorine, since we do not have 2 moles of sodium the chlorine is in excess.
Step 4
Now that we have identified the sodium as the limiting reagent, we are able to establish the expected yield. Looking back at the original equation, the number of moles of Na is the same as the number of moles of NaCl. Therefore if we have 1.05... moles of Na, we would theoretically expect to get 1.05... moles of NaCl.
Therefore the expected yield in grams would be (mass of one mole) x (number of moles we have). The expected yield is therefore 1.05... x (23+35.5) = 61.6g
Step 5
Now we have the expected yield we can work out the % yield.
% yield = mass we got/ expected mass x 100
= 26.4/61.6 *100 = 42.9%
Hope that helped 
vickie89uk
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 17022016 15:25
(Original post by samb1234)
Well done, and no worries, most of the questions they ask will just be slight variations on that type of question e.g. have concs of acids or volumes of gases instead of solids that kind of thing. Best of luck with everything, let me know if I can help with anything else 
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 17022016 15:31
(Original post by vickie89uk)
Am I correct in thinking that this is probably the hardest part of chemistry to get right?!
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