In the reaction #B_2H_6(g) + 3O_2(g) -> B_2O_3(s) + 3H_2O(g), ΔH = -2035 kJ#, how much heat is released when a mixture of #4.65# #g# #B_2H_6# and #2.94# #g# #O_2# is burned?

The most important thing to realize here is that you must first determine whether or not you're dealing with a limiting reagent.

This is important because it tells you the number of moles of diborane, #"B"_2"H"_6#, that actually take part in the reaction.

So, start by taking a look at the balanced chemical equation for this reaction

#"B"_2"H"_text(6(g]) + color(purple)(3)"O"_text(2(g]) -> "B"_2"O"_text(3(s]) + 3"H"_2"O"_text((g])#

The #1:color(purple)(3)# mole ratio that exists between diborane and oxygen gas tells you tha the reaction consumes #color(purple)(3)# moles of oxygen gas for every #1# mole of diborane that takes part in the reaction.

So, use the molar masses of diborane and oxygen gas to calculate how many moles of each you have in your samples

#4.56 color(red)(cancel(color(black)("g"))) * ("1 mole B"_2"H"_6)/(27.67color(red)(cancel(color(black)("g")))) = "0.1648 moles B"_2"H"_6#

and

#2.94 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(15.9994color(red)(cancel(color(black)("g")))) = "0.1838 moles O"_2#

Notice that you don't have enough oxygen gas to allow for all the moles of diborane to take part in the reaction #-># oxygen acts as a limiting reagent.

This means that the oxygen will be consumed before all the moles of dibirane are consumed. More specifically, only

#0.1838 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole B"_2"H"_6)/(color(purple)(3)color(red)(cancel(color(black)("moles O"_2)))) = "0.06127 moles B"_2"H"_6#

will take part in the reaction, the rest will be in excess.

Now, according to the thermochemical equation, when one mole of diborane reacts with three moles of oxygen gas, one mole of boron trioxide, #"B"_2"O"_3# and three moles of water are produced, and #"2035 kJ"# of heat are released.

In your case, #0.06127# moles of diborane will react with your sample of oxygen gas to give off

#0.06127 color(red)(cancel(color(black)("moles B"_2"H"_6))) * "2035 kJ"/(1color(red)(cancel(color(black)("mole B"_2"H"_6)))) = "124.68 kJ"#

Rounded to three sig figs, the answer will be

#q = color(green)("125 kJ") -># heat given off