Question

Calculate the heat when 128.4 mL of 0.500 M HCl is mixed with 300.0 mL of 0.495 M Ba(OH)2. Assuming that the temperature of both solutions was initially 25.0°C and that the final mixture has a mass of 428.4 g and a specific heat capacity of 4.18 J/°C · g, calculate the final temperature of the mixture.

Answer 1

Step-by-step explanation:

According to the statement the following chemical reaction takes place as

`2HCL_(2) (_(aq)) + Ba(OH)_(2) (_(aq))` ⇒
`BaCl_(2)(_(aq))  + 2H_(2)O(_(l))` ΔH = - 118KJ

Now the first step is to find no of Moles,It can be seen from the given equation that 2 moles of HCL react with one mole of Ba
`(OH)_(2)`

first convert mL to L so that units cancel to give no of moles

Moles of HCl = 0.1284L x 0.500 mol/L = 0.0642 moles

Moles of Ba
`(OH)_(2)` = 0.3 x 0.495 mol/L = 0.1485 moles

It can bee seen easily that there is 2 : 1 so HCl is limiting reagent,therefore it will be 0.05 moles of HCl reacting with 0.025 moles of Ba(OH)2.

HCL moles are used to calculate heat release

0.0642 moles of HCL at 118KJ/2 mol HCL = 3.78KJ =3787.8 Joules

∴ dH = m C dT where C is the heat capacity, m is the mass T is the temperature and H is the heat

putting values

3787.8 = 428.4 g x 4.18
`(J)/(C^(°) )` x dT

dT = 2.1 C°

Final temperature = initial + change in temperature

Temperature= 25C° + 2.1C°

Temperature = 27.1C° Ans