Answer to Question #350663 in General Chemistry for Kell

Question

Solve the following thermochemical reactions:

2 HCl + Pb

PbCl₂ + H₂

H = -1861 J/mole

What is the enthalpy if there was 3.86g of PbCl₂?

Answer 1

2HCl + Pb = PbCl₂ + H₂ ∆H = -1861 J/mol

M (PbCl₂) = 278 g/mol

278 g PbCl₂ — ∆H = -1861 J

3.86 g PbCl₂ — ∆H = X

X = 3.86 * (-1861) / 278 = -25.84

Answer: ∆H = -25.84 J