Determine heat released in kJ

Question

2 HCl(aq) + 1 Ba(s) → 1 BaCl₂(aq) + 1 H₂(g) ΔH°_rx = -538.88 kJ

71.41 g of Ba(s) is added to 290 mL of 0.616 M of HCl(aq)·

Answer 1

2 HCl(aq) + 1 Ba(s) → 1 BaCl₂(aq) + 1 H₂(g) ΔH°_rx = -538.88 kJ

mol Ba present = 71.41 g x 1 mol/137 g = 0.521 mol

mol HCl present = 0.290 L x 0.616 mol/L = 0.179 mol

Limiting reactant is HCl b/c mol ratio of HCl : Ba is 2:1

0.179 mol HCl x -533.88 kj/2 mol = -47.7 kJ