please help with my chemistry homework

Question

A 2.50 g sample of powdered zinc is added to 100.0 mL of a 2.00 M aqueous solution of hydrobromic acid in a calorimeter. The total heat capacity of the calorimeter and solution is 448 J/K. The observed increase in temperature is 21.1 K at a constant pressure of one bar. Calculate the standard enthalpy of reaction using these data.

Zn(s)+2HBr(aq)⟶ZnBr2(aq)+H2(g)

Answer 1

Zn(s) + 2HBr(aq) ==> ZnBr(aq) + H₂(g)

21.1 degrees x 448 J/deg = 9453 J = heat produced by the reaction

moles Zn = 2.50 g x 1 mol Zn/65.38 g = 0.0382 moles Zn

moles HBr = 0.100 L x 2.00 mol/L = 0.2 moles HBr

Limiting reactant = Zn since mole ratio of Zn:HBr is 1:2

∆Hrxn = 9453 J/0.0382 mol = 247,460 J/mol = 247 kJ/mole