Question

When 4.51 g of CaCl2 dissolved in 50.00 mL of water in a coffee cup calorimeter, the temperature of the solution rose from 22.6°C to 25.8°C.

Specific heat of the solution is equal to the specific heat of water = 4.18 J/gºC.
Density of the solution is equal to the density of water = 1.00 g/mL.

What is qsolution?

What is qreaction ?

What is ÎHrxn in kJ/mol of CaCl2 ?

Answer 1

Step-by-step explanation:

The heat gained by the solution = q

`q=mc* (T_(final)-T_(initial))`

where,

q = heat gained = ?

c = specific heat of solution=
`4.18 J/^oC`

Mass of the solution(m) = mass of water + mass of calcium chloride

Mass of water = ?

Volume of water = 50.00 mL

Density of water = 1.00 g/mL

Mass = Density × Volume

m = 1.00 g/mL × 50.00 mL = 50.00 g

Mass of the solution (m) = 50.00 g + 4.51 g =54.51 g

`T_(final)` = final temperature =
`25.8 ^oC`

`T_(initial)` = initial temperature =
`22.6 ^oC`

Now put all the given values in the above formula, we get:

`q=54.51 g* 4.18 J/g^oC* (25.8-22.6)^oC`

`q=729.126 J`

The heat gained by the solution is 729.126 J.

Heat energy released during the reaction = q'

q' = -q ( law of conservation of energy)

q' = -729.126 J

The heat energy released during the reaction is -729.126 J.

Moles of calcium chloride, n =
`(4.51 g)/(111 g/mol)=0.04063 mol`

`\Delta H_(rxn)=(q')/(n)=(-729.126 J)/(0.04063 mol)=-17,945.23 J/mol= -17.945 kJ/mol`

The ΔH of the reaction is -17.945 kJ/mol.