J.R. S. answered 04/09/20
Ph.D. University Professor with 10+ years Tutoring Experience
q = heat = heat capacity x ∆T
q = 448 J/ºC x 21.1ºC = 9408 J .. this is for 2.50 g Zn reacting with 100.0 ml of 2.00 M HBr.
We need to now find the heat (q) for 1 mole of Zn reacting with 2 moles HBr, as written in the equation.
2.50 g Zn x 1 mol Zn/65.39 g = 0.03823 moles Zn
0.1 L x 2 mol/L = 0.2 mol HBr so HBr is in excess and Zn is limiting.
q = ∆Hrxn = 9408 J/0.03823 moles Zn = 246,089 J/mol = -246 kJ/mol (3 significant figures)
Mostafa M.
Heat is given off by the reaction so the answer should be a negative value.10/29/22
William M.
-246 kJ/mol10/13/22