Question

You react 100.0 mL 1.01 M HBr(aq) with 99.1 mL of 1.05 M KOH(aq) solution. How many moles are actually neutralized?

The density of each solution is 1.04 g/mL. Cp_cal = 7.0 J/^oC; C_s = 4.18 J/g·^oC   The DT = 6.3 ^oC. Calculate q and DH/mole.

Answer 1

Balanced equation : HBr (aq) + KOH (aq) KBr (aq) + H₂O (l)

Given : concentration of HBr = 1.01 M

volume of HBr = 100.0 mL = 0.1000 L

moles of HBr = (concentration of HBr) * (volume of HBr in Liter)

moles of HBr = (1.01 M) * (0.1000 L)

moles of HBr = 0.101 mol

Similarly, moles of KOH = 0.104 mol

HBr is the limiting reactant because moles of HBr is less than moles of KOH

Actual moles neutralized = moles of HBr = 0.101 mol

Total volume of solution = (volume of HBr) + (volume of KOH)

Total volume of solution = (100 mL) + (99.1 mL)

Total volume of solution = 199.1 mL

Total mass of solution = (Total volume of solution) * (density of solution)

Total mass of solution = (199.1 mL) * (1.04 g/mL)

Total mass of solution = 207.064 g

Heat gained by solution = (Total mass of solution) * (specific heat of solution) * (T)

Heat gained by solution = (207.064 g) * (4.18 J/g.^oC) * (6.3 ^oC)

Heat gained by solution = 5452.8 J

Heat gained by calorimeter = (C_p,cal) * (T)

Heat gained by calorimeter = (7.0 J/^oC) * (6.3 ^oC)

Heat gained by calorimeter = 44.1 J

Heat lost by reaction, q = -(Heat gained by calorimeter + Heat gained by solution)

q = -(5452.8 J + 44.1 J)

q = -5496.9 J

H = q / moles of HBr

H = (-5496.9 J) / (0.101 mol)

H = -54425 J/mol

H = -54.4 kJ/mol