Here in this example question it says if the 466 kilojoules of energy is released for every mole of Dybborne reacting with water, how much energy would be released when 150 grams of Dybborne is submerged in excess water? All right, so remember our balance equation is 1 mole of dye. Boring gas reacts with six moles of water as a liquid to produce 2 moles of boric acid +6 moles of hydrogen gas. And we're just told here that the amount of energy, which we're going to say is ΔH released means -466 kilojoules, right?

So this becomes a thermal chemical question, Thermochemical question, because we're dealing with stoichiometry mixed with the enthalpy of the reaction. So the way we set this up is we have 150 grams of Dybarrane. What we have to do first is we have to convert these grams into moles. 1 mole of Dybborne on top, grams of diborin on the bottom so they can cancel out. When you calculate the two borons and the six hydrogens involved, you get a overall molar mass of 27.668 grams. Cancel out for diborane.

Now we have moles of dye borne according to my balanced equation for everyone, mole of Dybborne. This is how much energy is released. OK, so here we're not doing a mole to mole comparison, we're doing a ΔH to mole comparison. Everything cancels out. So what we have left at the end is negative 2526.38 kilojoules. This has three sig figs. This has four sig figs. So we'd have three sig figs at the end. So -2520 or actually 2530 kilojoules at the end. This is how much heat would be released from this many grams of Dybborne.