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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.

Respuesta :

Answer:

[tex]\Delta H[/tex] will be 90054 J

Explanation:

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

Molar mass of [tex]CuCl_{2}[/tex] = 134.45 g/mol

So, 1.00 g of [tex]CuCl_{2}[/tex] = [tex]\frac{1.00}{134.45}mol[/tex] of [tex]CuCl_{2}[/tex] = 0.00744 mol of [tex]CuCl_{2}[/tex]

0.00744 mol of [tex]CuCl_{2}[/tex] produces 670 J of heat

So, 1 mol of [tex]CuCl_{2}[/tex] produces [tex]\frac{670}{0.00744}J[/tex] of heat or 90054 J of heat

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