Consider the reaction B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) ∆H = -2035 kJ How much heat is released when a mixture of 8.64 g B2H6 and 7.18 g O2 is burned?

Question

Consider the reaction

B₂H₆(g) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(g) ∆H = -2035 kJ

How much heat is released when a mixture of 8.64 g B₂H₆ and 7.18 g O₂ is burned?

Answer 1

B₂H₆(g) + 3 O₂(g) → B₂O₃(s) + 3 H₂O(g) ∆H = -2035 kJ

This tells us that for every 1 mol of B₂O₃ that is formed in this reaction, 2035 kJ of heat are liberated.

So, we need to find out how many moles of B₂O₃ have been formed under the given conditions.

To do this, we must first find the limiting reactant, between B₂H₆ and O₂. One easy way to do this is to divide the moles of each by their corresponding coefficient in the balanced equation. Thus...

B₂H₆: 8.64 g B₂H₆ x 1 mol/27.67 g = 0.312 mols (÷1 -> 0.312)

O2: 7.18 g O₂ x 1 mol / 32 g = 0.224 (÷3 -> 0.074)

O₂ is LIMITING (0.074 is less than 0.312)

Now, we use moles of O₂ to find moles of B₂O₃ formed:

0.224 mols O₂ x 1 mol B₂O₃ / 3 mols O₂ = 0.0747 mols B₂O₃ formed

Finally, we compute the amount of heat released:

0.0747 mols B₂O₃ x 2035 kJ / mol B₂O₃ = 152 kJ of heat released

You could also report this as -152 kJ (the negative sign means heat is released)