The heat of neutralization of HCl (aq) by NaOH (aq) is -55.84 Kj/mol H2O produced.?

The heat of neutralization of HCl (aq) by NaOH (aq) is -55.84 Kj/mol H2O produced.





If 50.00 ml of 1.10 M NaOH is added to 25.00 ml of 1.90 M HCl, with both solutions originally at 24.70 C, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02 g/ml and a specific heat of 3.98 j/g/C.)


Express your answer using four significant figures.

The heat of neutralization of HCl (aq) by NaOH (aq) is -55.84 Kj/mol H2O produced.?
50.00 ml of 1.10 M NaOH contains: 0.05000*1.10 mol = 0.0550 mol of NaOH


25.00 ml of 1.90 M HCl contains: 0.02500*1.90 mol = 0.0475 mol of HCl.


Hence HCl is the limiting reactant, and thus 0.0475 mol of H2O may be formed. The total heat to be released is:


(55.84 Kj/mol)*(0.0475 mol) = 2.65 kJ





Once 50.00 ml of 1.10 M NaOH is added to 25.00 ml of 1.90 M HCl, the total volume is 75.00ml, which is 75.00*1.02 g = 76.5g.


The temperature increase post-mixing is:


∆T = 2.65x10^3/(3.98*76.5) = 8.71K


Therefore the final solution temperature is: 33.41ºC.
Reply:allow me to ask what or where did the constant?? 2.65X10^3 came from? Report It