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    (Original post by surina16)
    Thanks but I'm confused now aha - is 84 not the RFM?
    Oops, meant to quote the other person who said that the RFM of 2 moles was 168, which doesn't really make sense since the definition of RFM is that it's the mass of one mole of the compound relative to 12g of carbon 12. Ignore me lol
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    (Original post by samb1234)
    Otherway round, it's 16.8/84, don't worry about it everyone finds some stuff difficult. so now you know how many moles of nahco3 there are, you know by ratio that there are half the number of moles of na2co3. Then you just need to convert that back into a mass you can use
    45
    Na X 2 =46
    C=12
    o=32
    90/2= 45
    As we have half
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    (Original post by vickie89uk)
    45
    Na X 2 =46
    C=12
    o=32
    90/2= 45
    As we have half
    3 oxygens, meaning the total mass of one mole would be 46+12+48 = 106. We had 16.8/84 moles of sodium hydrogen carbonate = 0.2 moles, so how many moles do we expect to produce of na2co3
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    (Original post by samb1234)
    3 oxygens, meaning the total mass of one mole would be 46+12+48 = 106. We had 16.8/84 moles of sodium hydrogen carbonate = 0.2 moles, so how many moles do we expect to produce of na2co3
    9.20/106 = 0.08679..:::../
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    NaHCO3 -> Na2CO3
    ( 2 -> 1 )
    16.8g -> 9.20g
    84 -> 106/2
    84 -> 53
    So now divide Mr's: 53/84 = 0.63095......
    And multiply that number by 16.8g to find the theoretical mass of Na2CO3
    16.8g -> 10.6g ( our theoretical mass )

    Now simply:
    (Actuall mass / theoretical mass) x 100
    9.20/10.6 x100 = 86.8%
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    (Original post by vickie89uk)
    9.20/106 = 0.08679..:::../
    We had 0.2 moles of nahco3, and from the ratio we know that for every 2 moles of nahco3 there is one mole of na2co3. Therefore the max yield would be 0.1 moles of naco3, so we then need to multiply the RFM (106) bu=y 0.1 in order to find the max yield in grams
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    (Original post by samb1234)
    We had 0.2 moles of nahco3, and from the ratio we know that for every 2 moles of nahco3 there is one mole of na2co3. Therefore the max yield would be 0.1 moles of naco3, so we then need to multiply the RFM (106) bu=y 0.1 in order to find the max yield in grams
    1,060
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    (Original post by vickie89uk)
    1,060
    10.6 yes, so then the % yield will just be the yield we actually got (9.2) divided by 10.6 times 100
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    (Original post by samb1234)
    Oops, meant to quote the other person who said that the RFM of 2 moles was 168, which doesn't really make sense since the definition of RFM is that it's the mass of one mole of the compound relative to 12g of carbon 12. Ignore me lol
    no problem
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    (Original post by samb1234)
    10.6 yes, so then the % yield will just be the yield we actually got (9.2) divided by 10.6 times 100
    Ah that makes sense so that's 86.79%

    So can I clarify the steps for this So I can practice.

    Look at product we start with and products we produce
    We see there's a H2O and that's only possible if we started with 2NaHCO3 for there to be two hydrogen as a product.

    We look at what we collect which is Na2CO3 sodium carbonate
    We work out the relative mass of 2NaHCO3 sodium hydrogenated and then divide what we had to start by the maximum amount we could have made 16.8/84 0.2moles
    Then do the same for the sodium carbonate NaHCO3 has a Ar of 106 we divide
    9.20/106 = 0.08679 * 100 86.79% of sodium carbonate we actually made

    Is that right
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    (Original post by vickie89uk)
    Ah that makes sense so that's 86.79%

    So can I clarify the steps for this So I can practice.

    Look at product we start with and products we produce
    We see there's a H2O and that's only possible if we started with 2NaHCO3 for there to be two hydrogen as a product.

    We look at what we collect which is Na2CO3 sodium carbonate
    We work out the relative mass of 2NaHCO3 sodium hydrogenated and then divide what we had to start by the maximum amount we could have made 16.8/84 0.2moles
    Then do the same for the sodium carbonate NaHCO3 has a Ar of 106 we divide
    9.20/106 = 0.08679 * 100 86.79% of sodium carbonate we actually made

    Is that right
    Almost. Actually the easiest way to know there's 2 nahco3 is that na2co3 has 2 sodiums so there has to be 2 nahco3. Your method for the first bit is correct up unti lthe moles up until 0.2. That 0.2 is not the number of moles we could create, it's how many moles of NaHco3 we had at the start. You then need to look at the mole ratios to work out how many moles of the product that would give, so in this case 0.2 moles of NaHCO3 reacting would give 0.1 moles of Na2co3. You would then multiply the maximum possible number of moles of the product (in this case 0.1) by the mass of the product, so 0.1 x 106 = 10.6g max yield. You then divide 9.2/10.6 and x100
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    (Original post by samb1234)
    Almost. Actually the easiest way to know there's 2 nahco3 is that na2co3 has 2 sodiums so there has to be 2 nahco3. Your method for the first bit is correct up unti lthe moles up until 0.2. That 0.2 is not the number of moles we could create, it's how many moles of NaHco3 we had at the start. You then need to look at the mole ratios to work out how many moles of the product that would give, so in this case 0.2 moles of NaHCO3 reacting would give 0.1 moles of Na2co3. You would then multiply the maximum possible number of moles of the product (in this case 0.1) by the mass of the product, so 0.1 x 106 = 10.6g max yield. You then divide 9.2/10.6 and x100
    Is that because sodium is diatomic?! how do we know this (so in this case 0.2 moles of NaHCO3 reacting would give 0.1 moles of Na2co3) I'm sorry I just want to make sure I completely understand so you don't have to explain it to me again tomorrow... I think in understanding it I just need to clarify the method and how go through the process.... Thank you for your help
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    (Original post by vickie89uk)
    Is that because sodium is diatomic?! how do we know this (so in this case 0.2 moles of NaHCO3 reacting would give 0.1 moles of Na2co3) I'm sorry I just want to make sure I completely understand so you don't have to explain it to me again tomorrow... I think in understanding it I just need to clarify the method and how go through the process.... Thank you for your help
    Go back to the original balanced equation, that tell us for every 1 mole of na2co3 there is there are 2 moles of nahco3. Or to put it another way, 2moles of nahco3 react to form one mole of na2co3, so if we have a 10th of that amount (0.2 moles) 0.2 moles must react to form 0.1 moles
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    (Original post by samb1234)
    Go back to the original balanced equation, that tell us for every 1 mole of na2co3 there is there are 2 moles of nahco3. Or to put it another way, 2moles of nahco3 react to form one mole of na2co3, so if we have a 10th of that amount (0.2 moles) 0.2 moles must react to form 0.1 moles
    But we balanced the equation by realising that H must be in the original compound twice... I don't understand fully how that tells us for 1 mile of na2co3 we have 2 miles of nahco3.... Sorry
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    (Original post by vickie89uk)
    But we balanced the equation by realising that H must be in the original compound twice... I don't understand fully how that tells us for 1 mile of na2co3 we have 2 miles of nahco3.... Sorry
    doesn't matter why we balanced the equation the important bit is that the equation is 2 NaHco3 -> 1 Na2co3 ...
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    (Original post by samb1234)
    Almost. Actually the easiest way to know there's 2 nahco3 is that na2co3 has 2 sodiums so there has to be 2 nahco3. Your method for the first bit is correct up unti lthe moles up until 0.2. That 0.2 is not the number of moles we could create, it's how many moles of NaHco3 we had at the start. You then need to look at the mole ratios to work out how many moles of the product that would give, so in this case 0.2 moles of NaHCO3 reacting would give 0.1 moles of Na2co3. You would then multiply the maximum possible number of moles of the product (in this case 0.1) by the mass of the product, so 0.1 x 106 = 10.6g max yield. You then divide 9.2/10.6 and x100
    Ah so sodium always comes in pairs so there must be 2Na..... We started with 0.2 miles from the sodium hydrogenated and we then half that to 0.1 because we had two quantities I'm not sure entirely why we half it but 0.1 then we sodium carbonate has 106g and we collected 9.2
    0.1 how I don't know X 106= 10.60
    9.2/10.6 = 0.8679 x100= 86.79%
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    (Original post by samb1234)
    doesn't matter why we balanced the equation the important bit is that the equation is 2 NaHco3 -> 1 Na2co3 ...
    Yeah but I don't know how I figure I have a 2:1 ratio that's the only bit I don't understand
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    (Original post by vickie89uk)
    Yeah but I don't know how I figure I have a 2:1 ratio that's the only bit I don't understand
    because in order to balance the equation you had to have 2 nahco3
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    (Original post by samb1234)
    because in order to balance the equation you had to have 2 nahco3
    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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    This topic is for GSCE right?
 
 
 
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