Question

A chemist mixes 1.00 g CuCl2 with an excess of (NH4)2HPO4 in dilute aqueous solution . He measures the evolution of 670 J of heat as the two substances react to give Cu3(PO4)2(s). Compute the ΔH that would result from the reaction of 1.00 mo! CuCl2 with an excess of (NH4)2HPO4.

Answer 1

`\Delta H` will be 90054 J

Step-by-step explanation:

Number of moles = (mass)/(molar mass)

Molar mass of
`CuCl_(2)` = 134.45 g/mol

So, 1.00 g of
`CuCl_(2)` =
`(1.00)/(134.45)mol` of
`CuCl_(2)` = 0.00744 mol of
`CuCl_(2)`

0.00744 mol of
`CuCl_(2)` produces 670 J of heat

So, 1 mol of
`CuCl_(2)` produces
`(670)/(0.00744)J` of heat or 90054 J of heat