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Can someone explain in detail the steps to do thisproblem? A quantity of 4.00 x

ID: 676239 • Letter: C

Question

Can someone explain in detail 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? (Use heat of neutralization = -56.2 kJ/mol) Can someone explain in detail 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? (Use heat of neutralization = -56.2 kJ/mol)

Explanation / Answer

Balanced chemical equation :                                          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