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Help with Theoretical Yield |
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| Feb24-11, 12:31 AM | #1 |
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Help with Theoretical Yield
1. The problem statement, all variables and given/known data
Here's the measurements from my expirement. (reaction 1) 0.548 g Cu (s) 4.0 mL HNO3 100 mL DI water (reaction 2) 30 mL 3.0 M NAOH (reaction 3) nothing was added. (reaction 4) 200 mL of DI water 15 mL of 6.0 M HCl (reaction 5) 2.095 g Zn 10 mL 6M HCl 5 mL methanol 5 mL DI water 40.767 g mass of evaporating dish 41.238 g mass of evaporating dish with solid 2. Relevant equations Here are the reactions: reaction 1: theoretical mass of copper (II) nitrate formed Cu+4HNO3-->Cu(NO3)2+2NO2+2H2O reactoin 2: theoretical mass of copper (II) hydroxide formed Cu(NO3)2+2NaOH-->Cu(OH)2+2NaNO3 reaction 3: theoretical mass of copper (II) oxide formed Cu(OH)2-->CuO+H2O reaction 4: theoretical mass of copper (II) sulfate formed CuO+H2SO4-->CuSO4+H2O reaction 5: theoretical mass of copper formed Zn+CuSO4-->ZnSO4+Cu 3. The attempt at a solution I have no clue where to even start. my teacher said some of the measurements from the lab were needed in the calculations, but i dont know which measurements to use for each reaction. |
| Feb24-11, 04:05 AM | #2 |
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First of all - what are you expected to calculate?
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| Feb24-11, 01:39 PM | #3 |
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what I need to calculate (some of the measurements from the lab are used in the calculations):
the theorecital mass of copper (II) nitrate formed in reaction 1 the theoretical mass of copper (II) hydroxide formed in reaction 2 the theoretical mass of copper (II) oxide formed in reaction 3 the theoretical mass of copper formed and the % yield of the overall reaction im not asking you to do all those problems for me. but, i have no idea where to start. i need somebody to help show me how to do the first one, then i could do the rest on my own. |
| Feb24-11, 01:49 PM | #4 |
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Help with Theoretical YieldDo you know what limiting reagent is? |
| Feb24-11, 02:00 PM | #5 |
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all of the reaction questions are apart of my assignment, so i have to do them all.
i think the limiting reactant is the reactant that limits the amount of product. |
| Feb24-11, 02:24 PM | #6 |
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http://www.chembuddy.com/?left=balan...iting-reagents Start with the first reaction - what is a limiting reagent, copper, or nitric acid? Hm, I see you don't have concentration of nitric acid. If so, assume there was an excess of nitric acid. |
| Feb24-11, 04:19 PM | #7 |
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i tried working it out. is this right?
0.548g Cu*(1 mol Cu/63.55g Cu)=0.009g mol Cu 0.009 mol Cu*(1 mol Cu(NO3)2/1 mol Cu)=0.009 mol Cu(NO3)2 0.009 mol Cu(NO3)2*(125.56g Cu(NO3)2/1 mol Cu(NO3)2=1.08g Cu(NO3)2 |
| Feb24-11, 04:32 PM | #8 |
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0.009 seems to be rounded down way too far. You should use at least the same number of significant figures as you have in the copper mass. |
| Feb24-11, 04:35 PM | #9 |
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sorry i didnt mean to stick that g in there. what should i have done differently besides rounding?
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| Feb24-11, 05:00 PM | #10 |
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Other than things I listed what you did looks OK. Try the next step.
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| Feb24-11, 05:02 PM | #11 |
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is there more i need to do for the first reaction? and would i do the second reaction the same way?
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| Feb24-11, 05:08 PM | #12 |
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First reaction is OK. In general you may assume all other reactions can be done the same way, although - to be on the safe side - you should check each time if added reagents are in excess.
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| Feb24-11, 05:17 PM | #13 |
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would i use the answer from the first reaction to calculate the second reaction?
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| Feb24-11, 05:47 PM | #14 |
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i dont know if its right, but i used the answer from the first reaction to calculate the second reaction since all these reactions are connected.
1.08g Cu(NO3)2*(1 mol Cu(NO3)2/187.57g Cu(NO3)2)=0.00576 mol Cu(NO3)2 0.00576 mol Cu(NO3)2*(1 mol Cu(OH)2/1 mol Cu(NO3)2)=0.00576 mol Cu(OH)2 0.00576 mol Cu(OH)2*(97.566g Cu(OH)2/1 mol Cu(OH)2=0.562g Cu(OH)2 |
| Feb25-11, 02:47 AM | #15 |
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What is molar mass of copper nitrate? Other than that you are on the right track. Question: do you have to calculate number of moles of each substance each time? Or are they somehow dependent? |
| Feb25-11, 05:05 PM | #16 |
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yeah i realized my math was wrong on the first reaction. i fixed it. and my teacher said we are suppose to work in that 4.0 mL of HNO3. its suppose to be turned into mols. how would i do that?
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| Feb25-11, 05:57 PM | #17 |
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You can't without concentration. Perhaps it was just a stock concentrated solution, that would mean around 68% w/w. You have to check the solution density.
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