When a 3.76-g sample of solid iron(III) chloride dissolves in 47.9 g of water in

Question

When a 3.76-g sample of solid iron(III) chloride dissolves in 47.9 g of water in a coffee-cup calorimeter (see above figure) the temperature rises from 24.00 oC to 40.03 oC. Calculate H in kJ/mol FeCl3 for the solution process.FeCl3(s)  Fe3+(aq) + 3 Cl-(aq)

The specific heat of water is 4.18 J/g-K. When a 3.76-g sample of solid iron(III) chloride dissolves in 47.9 g of water in a coffee-cup calorimeter (see above figure) the temperature rises from 24.00 oC to 40.03 oC. Calculate H in kJ/mol FeCl3 for the solution process.FeCl3(s)  Fe3+(aq) + 3 Cl-(aq)

The specific heat of water is 4.18 J/g-K. When a 3.76-g sample of solid iron(III) chloride dissolves in 47.9 g of water in a coffee-cup calorimeter (see above figure) the temperature rises from 24.00 oC to 40.03 oC. Calculate H in kJ/mol FeCl3 for the solution process.FeCl3(s)  Fe3+(aq) + 3 Cl-(aq)

The specific heat of water is 4.18 J/g-K. When a 3.76-g sample of solid iron(III) chloride dissolves in 47.9 g of water in a coffee-cup calorimeter (see above figure) the temperature rises from 24.00 oC to 40.03 oC. Calculate H in kJ/mol FeCl3 for the solution process.FeCl3(s)  Fe3+(aq) + 3 Cl-(aq)

The specific heat of water is 4.18 J/g-K. When a 3.76-g sample of solid iron(III) chloride dissolves in 47.9 g of water in a coffee-cup calorimeter (see above figure) the temperature rises from 24.00 oC to 40.03 oC. Calculate H in kJ/mol FeCl3 for the solution process.FeCl3(s)  Fe3+(aq) + 3 Cl-(aq)

The specific heat of water is 4.18 J/g-K.

Explanation / Answer

heat released

Q =ms delta T

= (3.76 + 47.9 ) (4.18 )(40.03-24 )

=3461 J

= 3.46 KJ

-------------------------

moles of FeCl3

= (3.76/162.2)

=0.02318

---------------

delta H

= 3.46/ 0.02318

= -146.25 KJ <---------------answer

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