Can someone explain in a lot of details the steps to do thisproblem? A quantity of 4.00 x 102 mL of 0.600MHNO3 is mixed with 4.00 x 102 mL of 0.300MBa(OH)2 in a constant-pressure calorimeter of negligibleheat capacity. The initial temperature of both solutions is thesame at 18.46oC. What is the final temperature of thesolution? It says to use the result in another example for thecalculation. The result in that example is: The heat of neutralization = -56.2 kJ/mol. Can someone explain in a lot of details the steps to do thisproblem? A quantity of 4.00 x 102 mL of 0.600MHNO3 is mixed with 4.00 x 102 mL of 0.300MBa(OH)2 in a constant-pressure calorimeter of negligibleheat capacity. The initial temperature of both solutions is thesame at 18.46oC. What is the final temperature of thesolution? It says to use the result in another example for thecalculation. The result in that example is: The heat of neutralization = -56.2 kJ/mol.
Explanation / Answer
Ba(OH)2 + 2HNO3 ---------- >Ba(NO3)2 + 2 H2O 1 mol of barium hydroxide reacts with 2 mols of nitric acid. Number of mols of Ba(OH)2 = 0.4 Lx 0.300 M = 0.12 mols Number of mols of HNO3 = 0.4 L x 0.600 M =0.24 mols 0.12 mols of barium hydroxide neutralised by 0.24 mols ofnitric acid. For 1 mol heat of neutralization = -56.2 kJ For 0.24 mols of acid heat of neutralisation is -56.2kJ x 0.24 mols = -13.488 x 103 qsoln = m x s x T =800 g x 4.184 J / 0C g x ( T -18.46oC ) ( since density of water 1.0 / mL) q rxn = - q soln - 800 g x 4.184 J / 0C g x (T - 18.46oC ) = -13.488 x 103 J ( T - 18.46oC ) = -13.488 x 103 J/ - 800 g x 4.184 J / 0C g = 4.020C T = ( 4.02 + 18.46) 0C = 22.48 0C