Question

Part B Heat of Reaction for HCl(aq) + NaOH (aq) NaCl (aq) + so H20 (1) Volume of 1.0 M HCl (ml) so 100 23.3 Volume of 1.0 M NaOH (ml) Mass of Solution (8) use 1.00 g/ml Ti solution (temp of HCl just before reaction) T solution (highest temp obtained) Ti calorimeter (same as Ti solution, above) Te calorimeter (highest temp obtained) 100 23.3 29.8 23.3 29.8 23.3 - 29.8 Calculate q solution a solution = (mass of solution) x (4.184 / q°C) x T solution - T solution) 6.5 q solution (100) (4.184) (29.8-23.3)) 2719.4 q solution (5) 2719,6 2219,6 Calculate q calorimeter a calorimeter = (215/°C) x (T calorimeter - Ticalorimeter) qual-(21) (2918-23.3). = 134.5 q calorimeter (1) 136.5 1316.5 Calculate q reaction q calorimeter + solution + a reaction = 0 - 136.5-2719.6 = greaction from the first law of thermodynamics reaction (1) -2856.1

i need help on the second page:

"calculate how many mols of HCL you used based on the volume and molarity of the HCL"

and

"enthalpy of reaction" parts

Answer 1

For calculating the Moles the formula of molarity can be consisdered ccording to it

Molarity is dfined as the moles present per 1000 ml

The Molarity of the solution is 0.1 M

50 mL of solution contain (1/1000 ) * 50 = 5 * 10^-2

number of Moles of HCl are 5 * 10^-3 in both Cases

Now for calculating $\Delta$ H , We use the following relation that is

$\Delta$H = J / number of moles

J has been calculated as -2854.1 joules/ mole in both cases which is equivalent to -2.854 KJ / mole

  $\Delta$​​​​​​​H (J/mole)= -2854/5 * 10^-2 = -57082 J / mole

  $\Delta$​​​​​​​H ( KJ/ mole) = -57082 j / mole / 1000 = -57.082 KJ/mol

For both the cases the energy of Neutralization is -57.082 KJ/mol in both the cases

The average energy of neutralization is -57.082 KJ/mol