In this thermochemical scenario, we are tasked with determining the energy released when 150 grams of diborane (B₂H₆) reacts with excess water. The balanced chemical equation for this reaction indicates that 1 mole of diborane reacts with 6 moles of water to produce 2 moles of boric acid and 6 moles of hydrogen gas. The enthalpy change (ΔH) for this reaction is given as -466 kilojoules per mole, indicating that this amount of energy is released during the reaction.

To find the total energy released, we first need to convert the mass of diborane into moles. The molar mass of diborane can be calculated by adding the atomic masses of its constituent elements: 2 boron atoms (B) and 6 hydrogen atoms (H). The molar mass is calculated as follows:

\[\text{Molar mass of B}_2\text{H}_6 = (2 \times 10.81 \, \text{g/mol}) + (6 \times 1.008 \, \text{g/mol}) = 27.668 \, \text{g/mol}\]

Next, we convert 150 grams of diborane to moles using the formula:

\[\text{Moles of B}_2\text{H}_6 = \frac{150 \, \text{g}}{27.668 \, \text{g/mol}} \approx 5.42 \, \text{moles}\]

Now, we can calculate the total energy released by multiplying the number of moles of diborane by the energy released per mole:

\[\text{Energy released} = \text{Moles of B}_2\text{H}_6 \times \Delta H = 5.42 \, \text{moles} \times (-466 \, \text{kJ/mol}) \approx -2530.12 \, \text{kJ}\]

Considering significant figures, we round the final answer to three significant figures, resulting in an energy release of approximately -2530 kilojoules. This value represents the heat released when 150 grams of diborane reacts with excess water.