In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0
ID: 823585 • Letter: I
Question
In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.68
In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.68
In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.68
In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.68
In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.68 In a constant-pressure calorimeter, 70.0 mL of 0.310 M Ba(OH)2 was added to 70.0 mL of 0.620 M HCl. The reaction caused the temperature of the solution to rise from 21.68
Explanation / Answer
Q = mc dT = 140 ml x 4.186 j/goC x (25.9-21.68) = 2473.09 J or 2.473 kJ
HCl was in excess and it is so all the Ba(OH)2 is used up.
no of moles of Ba(OH)2 = 0.070*0.31 = 0.0217
So you have 0.01217 mol of Ba(OH)2 used up. That will produce 2 x 0.0217 mol of water or 0.0434 mol of water.
So the kJ / mol of water comes to 2.473 kJ / 0.0363 mol of water to get: 56.981 kJ/mol of water.