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

Answer:

$\Delta H$ will be 90054 J

Explanation:

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

Molar mass of $CuCl_{2}$ = 134.45 g/mol

So, 1.00 g of $CuCl_{2}$ = $\frac{1.00}{134.45}mol$ 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 $\frac{670}{0.00744}J$ of heat or 90054 J of heat