Question

Calculate the heat produced when 0.170 moles of boric acid, H₃BO₃ is produced.

B₂H₆ (g) + 6H₂O(l) → 2H₃BO₃(s) + 6H₂(g) + 184 kJ

Answer 1

As can be seen from the equation that for formation of 2 moles of boric acid it produces heat = + 184 kJ so, for 0.170 moles of boric acid (H₃BO₃) production $= \frac{184 \times 0.170}{ 2} = + 15.64 \; kJ$ of heat released