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    (Original post by samb1234)
    I can't really answer that because how difficult something is is completely subjective, so something that to you might make sense instantly might be someone elses hardest part of chemistry
    I see your point thank you again so much I feel confident I'll be ok if I keep practising
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    (Original post by vickie89uk)
    I see your point thank you again so much I feel confident I'll be ok if I keep practising
    No problem, if i can help in any other way just let me know and I'll try and get back to you
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    (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
    Just a few questions now I have gotten home and done this twice..... Your explanations and steps sequence is really fantastic I just have a few queries about how some parts works1)How do I know that sodium is a +1 ions and clorine is a 1- ion?! Is this because group 1 is +1 and group 7 is -1 or something.... Not a mention of anything like that in the book aqa endorsed and reccomendations nor in the Edexcel one.... I may be looking in the wrong place mind... 2)Would I work out the moles of chlorine if not given to me in the same way we worked out that 24.2/23=1.05 I'm assuming they'd give you a weight of some kind!!3)Do we need 1 mole of chlorine for every 2 moles sodium is that because we have a 2 Infront of Na or because chlorine is diatomic why do we have more sodium than chlorine ... I don't see how the equation represents this maybe I'm over analysing it.. I don't understand how cl is in excess as we have one mole of chlorine and 1.05 moles of sodium. This confuses me greatly... I don't know still how I figure that out.... Everything else in this part from the limiting reagent part is the same steps for yields isn't it?!
    3) 1.05 X 58.5 (23+35.5) = 61.48 I'm assuming I've buggered up somewhere Again thank you.... I've written an staircase of tasks on that so I know I first must balance and then find masses etc... Thanks again
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    (Original post by vickie89uk)
    Just a few questions now I have gotten home and done this twice..... Your explanations and steps sequence is really fantastic I just have a few queries about how some parts works1)How do I know that sodium is a +1 ions and clorine is a 1- ion?! Is this because group 1 is +1 and group 7 is -1 or something.... Not a mention of anything like that in the book aqa endorsed and reccomendations nor in the Edexcel one.... I may be looking in the wrong place mind... 2)Would I work out the moles of chlorine if not given to me in the same way we worked out that 24.2/23=1.05 I'm assuming they'd give you a weight of some kind!!3)Do we need 1 mole of chlorine for every 2 moles sodium is that because we have a 2 Infront of Na or because chlorine is diatomic why do we have more sodium than chlorine ... I don't see how the equation represents this maybe I'm over analysing it.. I don't understand how cl is in excess as we have one mole of chlorine and 1.05 moles of sodium. This confuses me greatly... I don't know still how I figure that out.... Everything else in this part from the limiting reagent part is the same steps for yields isn't it?!
    3) 1.05 X 58.5 (23+35.5) = 61.48 I'm assuming I've buggered up somewhere Again thank you.... I've written an staircase of tasks on that so I know I first must balance and then find masses etc... Thanks again
    Maybe it would be in the electronic structure part assuming you have got that far, basically at gcse level you just need to know that ions tend to have full outer shells of electrons, so group 7 atoms have 7 so gain 1 electron to get a full shell, group 1 has 1 so loses 1 to get a full outer shell. Yes, you would work out the moles of chlorine in the same way, or potentially use the fact that one mole of a gas at room temp and pressure occupies 24dm3. We know that we need 2 sodium for every chlorine from the 2 in front of soidum, whether it's diatomic or not makes no difference.

    Imagine the equation like a recipe. We take 2 sodiums and react them with one chlorine molecule, giving us 2 moles of NaCl. If we have 1 mole of chlorine and 1.05 moles of Na, then we have enough chlorine to make 2 moles of Nacl. However to make two moles of NaCl we would also need 2 moles of Na, and we only have 1.05 moles so therefore the sodium is the limiting reagent and chlorine is in excess. Regarding the 1.05 x58.5, you're right you've just used the rounded value of 1.05 rather than 1.055etc or whatever it was
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    (Original post by samb1234)
    Maybe it would be in the electronic structure part assuming you have got that far, basically at gcse level you just need to know that ions tend to have full outer shells of electrons, so group 7 atoms have 7 so gain 1 electron to get a full shell, group 1 has 1 so loses 1 to get a full outer shell. Yes, you would work out the moles of chlorine in the same way, or potentially use the fact that one mole of a gas at room temp and pressure occupies 24dm3. We know that we need 2 sodium for every chlorine from the 2 in front of soidum, whether it's diatomic or not makes no difference.

    Imagine the equation like a recipe. We take 2 sodiums and react them with one chlorine molecule, giving us 2 moles of NaCl. If we have 1 mole of chlorine and 1.05 moles of Na, then we have enough chlorine to make 2 moles of Nacl. However to make two moles of NaCl we would also need 2 moles of Na, and we only have 1.05 moles so therefore the sodium is the limiting reagent and chlorine is in excess. Regarding the 1.05 x58.5, you're right you've just used the rounded value of 1.05 rather than 1.055etc or whatever it was
    Ah I see... What's dm3 thankyou
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    decimetres cubed
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    (Original post by samb1234)
    decimetres cubed
    Thank you I haven't come across that yet either... Might have to take my exams in January I'm not sure November sits are availableAttachment 505225505227Name:  image.jpg
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    My friend has this book her daughter used for chemistry from cgp do you think this will help or do you think they aren't helpful.... Screenshot a couple of pages

    Thanks
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    (Original post by vickie89uk)
    Thank you I haven't come across that yet either... Might have to take my exams in January I'm not sure November sits are availableAttachment 505225505227Name:  image.jpg
Views: 44
Size:  514.2 KB

    My friend has this book her daughter used for chemistry from cgp do you think this will help or do you think they aren't helpful.... Screenshot a couple of pages

    Thanks
    The cgp guides were what i personally used for my gcse revision (although i already knew everything so they were just for recapping) but they were very concise and had the vast majority of the stuff i actualy needed to know whereas some of the text books tended to have a lot of irrelevant facts that you didn't really need
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    I found this website which explains most of chemistry for self teaching if anyone like me has a rubbish teacher.... http://chemistry.about.com/od/chemis...schoolchem.htm

    I seem to be able to understand most of the explanations in balancing equations yields etc
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    (Original post by samb1234)
    No problem, if i can help in any other way just let me know and I'll try and get back to you
    Still don't understand everything fully.... So I'm working overtime this week to pay to see a chemistry tutor next week. Hopefully I'll get it..... It can't be that hard but some are simple some don't need the entire mass numbers added together.... Some do.... I don't know when or when not to add fully.... Nightmare after a week I can't cope anymore the very idea of chemistry makes me feel sick
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    (Original post by samb1234)
    decimetres cubed
    Had a session with a tutor today £40 but I finally understand what moles means.... He said I did well thank you because a lot of what you told me finally set into place fully... He said my textbook is back to front and yielding and equations came before moles and that's why I was confused.... Thanks again
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    (Original post by vickie89uk)
    Had a session with a tutor today £40 but I finally understand what moles means.... He said I did well thank you because a lot of what you told me finally set into place fully... He said my textbook is back to front and yielding and equations came before moles and that's why I was confused.... Thanks again
    Ah i'm glad you were able to find someone to help you irl, and that the session went well and that you now get it. Best of luck for all of your exams!
 
 
 
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