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Homework Help: Chemistry

Posted by Hannah on Monday, February 13, 2012 at 7:18pm.

I had to perform an experiment using a coffee-cup calorimeter to determine the enthalpies of two seperate reactions. For part 1 we worked with magnesium metal, and magnesium oxide. Data was collected and now I need to find the q, heat transferred for both magnesium and magnesium oxide.

For magnesium the mass of rxn mixture was 24.371g and for magnesium oxide it was 25.036g.

I think that I am suppose to use the q=mc delta T equation and the rxn value but I am not sure quite how to do this. Do I just plug the rxn number into the equation along with the delta T value I got for both?



Chemistry - DrBob222, Monday, February 13, 2012 at 3:58pm
Yes you are to use q = mcdeltaT
What was the reaction and much of the other reagents did you use?



Chemistry - Hannah, Monday, February 13, 2012 at 5:04pm
0.150g of Mg was used and 0.250g MgO was used.For the first reaction we had to mix 25ml of HCl with the Mg and record the temp with a Labpro interface device. The second reaction we had to use the same thing except with MgO.

So for the mass of rxn mixture for Mg the value is 24.371g and delta T is 23.83degrees celsius. So I would do
q=(24.371)(23.83)I am not sure what c is?



Chemistry - Hannah, Monday, February 13, 2012 at 5:05pm
They give the specific heat capacity for water, is that what I use?



Chemistry - Hannah, Monday, February 13, 2012 at 5:16pm
So for Magnesium I did
q=(4.184 J/g X K)(24.371g)(296.83K) = 30267.23 J

Is this correct so far?



Chemistry - DrBob222, Monday, February 13, 2012 at 5:24pm
I am having a little trouble understanding exactly what you did and exactly what you want to calculate? Generally, these experiment are to determine q/g or q/mol for a substance, in this case q/g for Mg and q/g for MgO.
The other thing I'm having trouble with is you had 0.15 g Mg and added 25.00 mL HCl to it and the mass was not 25.00+0.15. Either I'm off base or you didn't add 25.00 mL HCl. Is that 24.371 g the total mass of the HCl added (24.221 g HCl + 0.15 g Mg = 24.371 g total) or is that something different. Same question for MgO. Next, did all of the Mg dissolve or was there some left at the end of the experiment?



Chemistry - Hannah, Monday, February 13, 2012 at 5:36pm
Im sorry it so confusing and there is alot of data. This whole experiment was split into two parts. I had to work in a group with two other people to do part 2 and for part 1 we had to get the data from another group so I just have some of their data. For part I guess the whole point was to determine the deltaH formation of magnesium oxide.

The data that I collected from the other group was mass of weigh dish, mass of weigh dish +sample, mass of sample, molecular weight of Mg and MgO, mass of styrofoam cups, mass of cup+ rxn mixture, mass of rxn mixture and then we created graphs on the computer and we got the Tinitial, TFinal, and delta T. This is all for magnesium and magnesium oxide. Now it asks to calculate the q for Mg and MgO. Then I need to calculate deltaH rxn for both, and then finally deltaH formation of MgO.

The three equations that were provided to determine the equation and value for delta H formation were:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g) Delta H1 = ?Kj

MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l) deltaH2=?Kj

H2(g) + 1/2(g)-> H20(l) delta H3 = -241.8kJ


Chemistry - Hannah, Monday, February 13, 2012 at 6:04pm
It also says to assume that none of the heat produced in the reaction is lost by the calorimeter to the rest of the surroundings. It says that the energy as heat lost by reaction will equal that gained by the solution which is represented by qreaction =-qsolution.

So for Mg I got 30267.23J. Is this suppose to be negative or positive?



Chemistry - Hannah, Monday, February 13, 2012 at 6:07pm
Then it says that to find deltaH rxn the equation is qrxn/mol substance.
So I use the 30267.23 / 0.152 to get my answer correct?



Chemistry - Hannah, Monday, February 13, 2012 at 6:11pm
Or should I use the # of moles that I calculated for Mg which was 6.25e-3.

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