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777888

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« on: January 13, 2005, 10:51:55 AM »
1. Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)
100.0mL of 1.0 M HCl(aq) is placed in a coffee cup. Ti=20.8C. 0.50g of Mg is added and Tf=44.8C. Determine the heat of reaction.

Work:
Energy absorbed by solution=100x4.18x24
(Can I ASSUME it has specific heat capacity of 4.18J/gC ?)
q=10.032kJ
Energy released by system=10.032kJ
(Is the question asking for molar enthalpy or enthalpy change?)
I need *delete me* Thank you!
« Last Edit: January 14, 2005, 10:05:29 PM by 777888 »

777888

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Re:heat of reaction
« Reply #1 on: January 13, 2005, 05:05:56 PM »
Do I have to do the limiting reagent thing and how?

Donaldson Tan

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Re:heat of reaction
« Reply #2 on: January 13, 2005, 09:21:45 PM »
yes. you gotta find use the "limiting agent" thingy because not everything has reacted. deltaH is per mole of reactant used up.
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777888

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Re:heat of reaction
« Reply #3 on: January 13, 2005, 11:14:12 PM »
yes. you gotta find use the "limiting agent" thingy because not everything has reacted. deltaH is per mole of reactant used up.
I found that Mg is limiting reagent...then are my calculations correct?
thanks

777888

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Re:calculations...heat of rxn
« Reply #4 on: January 14, 2005, 05:55:39 PM »
2. MgO(s)+2HCl(aq)->MgCl2(aq)+H2O(l)
100.0mL of 1.0mol/L HCl(aq) is placed in a coeffee cup(Initial temp.=20.4C). 1.00g MgO(s) is added and the final temp.=29.6C

Work:
Surroundings:
q=(100g)(4.18J/gC)(9.2C)=3.8456kJ
System:
MgO(s) + 2HCl(aq)  ->  MgCl2(aq)+H2O(l)
1.00g         0.1L
0.0248mol  0.1M
0.1mol
MgO(s) is limiting reagent.
Energy released=-3.8456kJ
Delta Hx = -3.8456kJ/0.02481mol
= -155.00kJ/mol MgO
= -1.6x10^2kJ/mol MgO

Can someone see if calculate it right? Just to make sure I get this stuff...Many thanks

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« Reply #5 on: January 14, 2005, 10:25:52 PM »
Looks good to me.  You can also check for yourself if you did it properly by looking up the delta Heat of formation for solid MgO, aqueous HCl, Aqueous MgCl2, and liquid water.  Then just take the sum of the products' heat of formation and subtract the the sum of the reactants' heat of formation.  That should give you the theoretical heat of reaction.
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777888

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« Reply #6 on: January 15, 2005, 12:02:24 AM »
Looks good to me.  You can also check for yourself if you did it properly by looking up the delta Heat of formation for solid MgO, aqueous HCl, Aqueous MgCl2, and liquid water.  Then just take the sum of the products' heat of formation and subtract the the sum of the reactants' heat of formation.  That should give you the theoretical heat of reaction.
When using your method to find "the theoretical heat of reaction", would that be in kJ or kJ/mol? If it is in kJ, how can I change it to kJ/mol and compare to my answer?

I have one more question, when it appears like below, what is the kJ/mol refering to? kJ/mol O2? kJ/mol H2O? Or...?
H2(g)+ 1/2 O2(g) -> H2O(l)    Delta Hx=-285.8kJ/mol

777888

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« Reply #7 on: January 15, 2005, 06:20:14 PM »
Looks good to me.  You can also check for yourself if you did it properly by looking up the delta Heat of formation for solid MgO, aqueous HCl, Aqueous MgCl2, and liquid water.  Then just take the sum of the products' heat of formation and subtract the the sum of the reactants' heat of formation.  That should give you the theoretical heat of reaction.
Hi judrg, my text book doesn't have standard enthalpies for Mg(s), HCl(aq), MgCl2(aq), and H2(g)! Where can I find them? Thank you!

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« Reply #8 on: January 16, 2005, 01:17:29 AM »
When using your method to find "the theoretical heat of reaction", would that be in kJ or kJ/mol? If it is in kJ, how can I change it to kJ/mol and compare to my answer?

I have one more question, when it appears like below, what is the kJ/mol refering to? kJ/mol O2? kJ/mol H2O? Or...?
H2(g)+ 1/2 O2(g) -> H2O(l)    Delta Hx=-285.8kJ/mol