Asked by Chris
What is the molar heat of reaction?
When 0.732 g of Ca metal is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 119C is observed.
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
The molar heat of reaction, H rxn , for the reaction of
Ca(s) + 2H+(aq) Ca2+(aq) + H2(g)
is ? kJ/mol
So, I believe i use the equation q=m*s*delta T
and i divide it by n (solute)?
I keep getting the wrong answer. Please help
When 0.732 g of Ca metal is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 119C is observed.
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
The molar heat of reaction, H rxn , for the reaction of
Ca(s) + 2H+(aq) Ca2+(aq) + H2(g)
is ? kJ/mol
So, I believe i use the equation q=m*s*delta T
and i divide it by n (solute)?
I keep getting the wrong answer. Please help
Answers
Answered by
drwls
Use the temperature rise, mass and specific heat of the final product to determine the heat release. Divide that by the number of moles of the reactant calcium. (I am assuming that Ca is the limiting reactant, and disappears; this should be checked)
Answered by
Chris
btw, i meant to write 11.9C not 119.
i work it like this:
200ml(1g/ml)(4.184J/C)(11.9C)
then divide by (40.078g/mol/.732g)
i get the wrong answer though
i work it like this:
200ml(1g/ml)(4.184J/C)(11.9C)
then divide by (40.078g/mol/.732g)
i get the wrong answer though
Answered by
DrBob222
Divide first by 0.732 to give H/gram, then multiply by 40.078 to obtain H/mol. That will give you the right answer.
Answered by
Chris
i get 545 but i'm not getting the right answer
Answered by
Chris
nvm its -545. not quite sure why its exothermic
Answered by
Chris
oh and thank you guys
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