50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calori
ID: 75130 • Letter: 5
Question
50.0 mL of a solution of HCl is combined with 100.0 mL of 1.05M NaOH in a calorimeter. The reaction mixture is initially at 22.4 degrees celsius, and the final temperature after reaction is 30.2 degrees celsius. What is the molarity of the HCl solution? You may assume that there is an excess of base (so that all the HCl has reacted), that the specific heat of the reaction mixture is 0.96 cal/g C, and that the density of the reaction mixture is 1.02 g/mL. The heat of neutralization of HCl and NaOH is 13.6 kcal/mole.Explanation / Answer
Total volume of the mixture = 50 mL + 100mL=150 mL density = 1.02 g/mL total mass = 150 mL * 1.02 g/mL = 153 g specific heat of the mixture = s = 0.96 cal / g.oC temperature change = T = 30.2oC -22.4oC = 7.8oC The heat released during the neutralization reaction is used towarm the mixture . Q = 153 g * 0.96 cal / g.oC *7.8oC = 1145.664 cal 1145.664 cal of heat is released during the reaction but the heatof neutralization per mole is 13.6 kcal. NaOH + HCl NaCl + H2O + heat 1 mol 1mol 1 mol Since NaOH is taken in excess HCl is the limiting reactant. HCl - heat of neutralization 1mol 13.6 x 103 cal ? 1145.664 cal Number of moles of HCl reacted = 0.08424 mol Volume of HCl = 500 mL = 0.05 L molarity of HCl = 0.08424 mol/ 0.05 L = 1.6848 M
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