Short answer: 15.6 kJ of heat is produced. Written as an enthalpy change for that amount of reaction, that is delta H = -15.6 kJ.
Step 1 - notice where the 184 kJ sits.
B2H6(g) + 6 H2O(l) -> 2 H3BO3(s) + 6 H2(g) + 184 kJ
The energy is written on the PRODUCT side, so heat comes out along with the products. The reaction is exothermic and delta H = -184 kJ, even though the number in the equation carries a plus sign. That plus sign is the single most common thing students misread here. Energy on the right means released; energy on the left means absorbed. The "+184 kJ" is not a positive delta H, it is 184 kJ of heat listed as a product.
Step 2 - attach the 184 kJ to the right amount.
The coefficient in front of H3BO3 is 2, so the 184 kJ goes with 2 moles of boric acid, not one:
184 kJ / 2 mol H3BO3 = 92.0 kJ per mole of H3BO3
Step 3 - scale it to 0.170 mol.
q = 0.170 mol H3BO3 x (184 kJ / 2 mol H3BO3) q = 0.170 x 92.0 q = 15.64 kJ -> 15.6 kJ
15.6 kJ of heat is produced.
Check it before you trust it. 0.170 mol is a small fraction of the 2 mol the equation is written for. 0.170 / 2 = 0.085, or 8.5 percent, so the answer has to be about 8.5 percent of 184 kJ, and 8.5 percent of 184 is roughly 15.6. That one-line estimate is worth doing every time, because both of the usual mistakes land far outside it.
The two wrong answers this problem produces.
31.3 kJ comes from using 184 kJ per mole of H3BO3 and ignoring the 2. It is exactly double the correct answer, and that factor of 2 is the fingerprint: if your result is double or half of a classmate's, one of you dropped a coefficient. Thermochemical energy belongs to the whole equation as written, never to a single species in it.
Anything built on 61.83 g/mol means you reached for the molar mass of boric acid. There are no grams anywhere in this problem. Moles were handed to you, and the conversion you actually need is moles-to-kilojoules, which is the coefficient ratio. Molar mass would only enter if the question had said 0.170 grams.
Significant figures. 0.170 has three and 184 has three, so the answer keeps three: 15.6 kJ. The 2 in the ratio is a stoichiometric coefficient, exact by definition, and it never limits the count. Writing 15.64 claims precision the data does not support.
The same equation answers three other questions, and it is worth writing all four out once. Each one is the same 184 kJ divided by a different coefficient:
per 1 mol B2H6 consumed: 184 kJ / 1 = 184 kJ per 1 mol H2O consumed: 184 kJ / 6 = 30.7 kJ per 1 mol H3BO3 formed: 184 kJ / 2 = 92.0 kJ per 1 mol H2 formed: 184 kJ / 6 = 30.7 kJ
If a later part of the assignment asks about grams of diborane or liters of hydrogen, convert to moles first and then use the matching line above. The conversion factor is never "184 kJ" on its own.
What this reaction actually is. It is the hydrolysis of diborane, and it is violently exothermic in practice. Diborane is pyrophoric, it ignites on contact with air, and moist air alone will start this reaction. That is worth knowing, because 184 kJ from one mole of a light gas is a great deal of energy in a very small package, which is exactly why the U.S. Air Force spent much of the 1950s trying to burn boranes as high-energy jet fuels. The effort was abandoned largely because the solid boron oxide residue fouled the engines. The chemistry in your homework is the same chemistry that ended that program.
