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You want to determine ΔH° for the reaction Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)....
You want to determine ΔH° for the reaction Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g). To do so, you first determine the heat capacity...
...of a calorimeter using the following reaction, whose ΔH is known: NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(I)
ΔH° = -57.32KJ
Calculate the heat capacity of the calorimeter from these data:
Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH
Initial T of both solutions: 16.9°C
Maximun T recorded during reaction: 30.4°C
Density of resulting NaCl solution: 1.04 g/mL
c of 1.00 M NaCl(aq) = 3.93 J/g*K
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The volume of the final solution is V =50+50 =100 mL with a density of 1.04 g/mL which means a total mass of solution
`m =rho*V =1.04*100 =104 g`
The balanced chemical equation of transformation is
`NaOH +HCl=> NaCl + H_2O`
Number of moles of NaOH entering is n(NaOH) = 2*50/1000 =0.1
Number of moles of HCl entering is n(HCl) =2*50/1000 =0.1 moles
For 0.1 mole of NaOH and 0.1 mole of HCl it results 0.1 mole of NaCl
The molecular mass of NaCl is M(NaCl) =23+35.5 =58.5 g
The molarity of the final solution is n =0.1 mole/V =0.1/0.1 = 1 mole/L
The heat released by the reaction (the enthalpy given in problem is in KJ/mole)
`Q =n*Delta(H) =0.1*57.32 =5.732 KJ`
The heat balance is
`m_(sol)*c_(sol)*Delta(T) +C_(cal)*Delta(T) =Q`
`C_(cal) = (5732-104*3.93*(30.4-16.9))/(30.4-16.9) = 15.87 J/K`
The heat capacity of the calorimeter is C =15.87 J/K
Posted by valentin68 on October 5, 2013 at 7:47 AM (Answer #1)
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