Question Details: A quantity of 4.00x10 2 mlof 0.600M HNO3 is mixed with4.00x10

Question

Question Details: A quantity of 4.00x102 mlof 0.600M HNO3  is mixed with4.00x10 2 mlof 0.300M Ba(OH)2 in a constant -pressure calorimeterof negligible heat capacity.the initial temperature of bothsolutions was the same 18.46C.what is the final temperature of thesolution ? Knowing that the heat of neutralization is: -56.2kJ/mol = Question Details: A quantity of 4.00x102 mlof 0.600M HNO3  is mixed with4.00x10 2 mlof 0.300M Ba(OH)2 in a constant -pressure calorimeterof negligible heat capacity.the initial temperature of bothsolutions was the same 18.46C.what is the final temperature of thesolution ? Knowing that the heat of neutralization is: -56.2kJ/mol = Question Details: A quantity of 4.00x102 mlof 0.600M HNO3  is mixed with4.00x10 2 mlof 0.300M Ba(OH)2 in a constant -pressure calorimeterof negligible heat capacity.the initial temperature of bothsolutions was the same 18.46C.what is the final temperature of thesolution ? Knowing that the heat of neutralization is: -56.2kJ/mol =

Explanation / Answer

For a neutralization reaction, 2 HNO3 (aq) + Ba(OH)2(aq) ---------> BaNO3 (aq) + 2 H2O(l)                   H = -56.2 kJ/mol   Since it was a constant pressure calorimetry atnegligible heat capacity, we can assume that qrxn = -Hneu Let us assme that the density of the solution is 1.0 g/mL, therefore - Total numbe of moles of HNO3 in the solution = 0.4 L *0.6mol/L = 0.24 mol total number of moles of Ba(OH)2 = 0.4 L *0.3 mol/L                                                      =0.12 mole thus for these moles, Hneu =-56.2 kJ/mol / 0.12 mole                                             = 468.33 kJ We know that,              q = msT         468.33 = (400 g +400 g) * 4.184 J/g .oC * (T - 18.46)      Final temperature = 18.59 oC   Let us assme that the density of the solution is 1.0 g/mL, therefore - Total numbe of moles of HNO3 in the solution = 0.4 L *0.6mol/L = 0.24 mol total number of moles of Ba(OH)2 = 0.4 L *0.3 mol/L                                                      =0.12 mole thus for these moles, Hneu =-56.2 kJ/mol / 0.12 mole                                             = 468.33 kJ We know that,              q = msT         468.33 = (400 g +400 g) * 4.184 J/g .oC * (T - 18.46)      Final temperature = 18.59 oC  

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