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| | From:  Albert1145 (Original Message) | Sent: 10/24/2008 6:02 AM |
Steve,
I don't know what equations to use for these two questions
1. How many grams of ethylene (C2H4) would have to be burned to produce 430.4 kJ of heat?
C2H4(g) + 3O2(g) -> 2CO2(g) + H2O(l) ΔHorxn=-1411 kJ
2. Ethanol fuel (C2H5OH) burns according to the equation below.
What is the value of qp (kJ) when 20.3 g of ethanol is burned?
C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l) Δ H= -1367 kJ |
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| | | From: ·Steve· | Sent: 10/24/2008 5:55 PM |
C2H4 (g) + 3 O2 (g) ––�?gt; 2 CO2 (g) + 2 H2O (l) ΔH° = �?411 kJ
What this balanced reaction says is this: One mole of ethylene reacts with three moles of oxygen to form two moles of carbon dioxide, two moles of water, and 1411 kJ of heat.
So, if one mole of ethylene produces 1411 kJ of heat, how many moles of ethylene will produce 430.4 kJ of heat?
You can do this like a conversion problem, "given" value X conversion factor(s) = "desired" value. The "given" value is 430.4 kJ. The "desired" value is the moles of ethylene. The conversion factor comes from the reaction,
"1411 kJ per 1 mol C2H4" or "1 mol C2H4 per 1411 kJ"
1411 kJ or 1 mol C2H4
mol C2H4 1411 kJ
We want to use the conversion factor that gives the correct units (moles of C2H4) in the end:
430.4 kJ X 1 mol C2H4 =
1 1411 kJ
kJ units cancel, leaving us with moles of C2H4. 
Convert moles of C2H4 to grams, and that's it!
The second problem is very similar.
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Thank you so much! I got three more questions though... 1. Car air bags operate by the explosive reaction of NaN3: 2NaN3(s) --> 2Na(s) + 3N2(g) What mass of NaN3 must be used to fill a 70L air bag at 20 celcius and 1 atm pressure? I keep on getting 189, could you please show me where I got wrong? 2. The following reaction takes place at 300K under a constant external pressure of 10^6 Pa. What is the work done if 20ml of NO2(g) is formed during the reaction? 2NO(g) + O2(g) --> 2NO2(g) I did work = -p(delta)nRT and got 2494200000!! I'm not sure where to do the unit conversion. Finally, I saw the annoucement that MSN groups will close in Feb 2009, does that mean no more Chemistry Corner? |
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| | | From: ·Steve· | Sent: 11/6/2008 1:58 AM |
Hi Albert! Yes, with MSN Groups closing, I'll either have to give up doing a chemistry help site, or move to another service. The MSN groups can be moved to the Multiply site, but it does not have the message board format like Groups does, so I am considering other options like Yuku or Aimoo that are better suited for this purpose. Well, it was fun while it lasted! There are some other good chemistry groups out there, so I'm also considering just joining one of those. Meanwhile, let's see what you've got here.... 1. First you need to calculate the moles of N2 in 70 L at 1 atm and 20°C. Use PV = nRT for that, to get n = 2.91 mol. Then, look at the reaction to see how many moles of NaN3 are needed. From the coefficients in the balanced reaction, we can see that moles of NaN3 will be 2/3 times the moles of N2. Finally, multiply moles of NaN3 by its molar mass (65.01 g/mol) to get grams of NaN3. I got 126 g. 2. The work w = –PoppΔV, where Popp, the opposing pressure, is 1000000 Pa. Since 1 atm = 101325 Pa, this is 9.86923 atm. Using atm for pressure and L for volume, –PoppΔV = �?9.86923 atm)(0.020 L) = �?.197385 L atm. We can covert to joules using the coversion 1 L atm = 101.325 J: �?.197385 L atm X 101.325 J / (L atm) = �?0 J.  |
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Sorry, I'm don't really understand question 2.
"–PoppΔV = �?9.86923 atm)(0.020 L) = �?.197385 L atm"
where the 0.02L come from. I though that is number of moles, and -20J isn't one of the options in the question.
I came across two acid and base equations last nigh, pretty similar but I can't do them. I think I have some problems with writing out chemical equation and the ICE table, could you help me with these two also?
10.3 mL 0.200 M nitric acid solution is added to 50.00 mL 0.100 M sodium hydroxide solution. What is the pH of the solution?
24.2 mL 0.200 M nitric acid solution is added to 50.00 mL 0.200 M ammonia solution. What is the pH of the solution? (Kb for ammonia 1.80 x 10-5)
Sorry for the trouble. |
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| | | From: ·Steve· | Sent: 11/7/2008 5:15 PM |
Hi Albert, let's look at the Popp question again. Popp is the opposing atmospheric pressure. When a gas forms in a reaction, it has to expand against the opposing atmospheric pressure. It takes work to do this, called "pressure-volume" work, w = –PoppΔV, where ΔV is the change in volume of the gases in the reaction. What I assumed incorrectly was that ΔV is 20 mL (the volume of NO2 formed) but I forgot that in this reaction, we are starting with gases too, which will be consumed as the NO2 product forms. Therefore, ΔV of this reaction will be the volume of the gaseous product minus the volume of the gaseous reactants. What was the original volume of the NO and O2 reactants? We can tell from the balanced reaction, because the volumes will be related to each other according to the coefficients in the reaction. 2 NO (g) + O2 (g) ––�?gt; 2 NO2 (g) Since the coefficients for NO and NO2 are the same, the moles of NO reacted and NO2 formed are the same. This also means that their volumes are the same. Since 20 mL of NO2 formed, we must have started with 20 mL of NO. Similarly, we can see from the reaction that the moles of O2 is 1/2 the moles of NO, and therefore half the volume also, which is 10 mL. So, the total volume of the reactants is 20 mL of NO plus 10 mL of O2 = 30 mL. Now we can get ΔV: ΔV = Vfinal �? Vinitial = Vproducts �? Vreactants = 20 mL �? 30 mL = �?0 mL. And now we can calculate the work: w = –PoppΔV = �?9.86923 atm)(�?.010 L) = +0.0986923 L atm Converting L atm to joules gives the work as +10 J. Hopefully, that is one of the answers!  Your other two questions involve the reaction of a strong acid with a strong base and a strong acid with a weak base. I'll look at those in the next post. |
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| | | From: ·Steve· | Sent: 11/7/2008 6:57 PM |
I moved the acid-base questions to a new thread here. |
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