Final answer:
Approximately -1758 kJ of heat will be transferred when 25.1 g of diborane reacts with 57.0 g of oxygen, as diborane is the limiting reagent and the stoichiometry of the reaction indicates that 1 mole of diborane releases 1940 kJ of energy.
Step-by-step explanation:
To determine how much heat will be transferred when 25.1 g of diborane (B₂H₆) reacts with 57.0 g of oxygen (O₂), according to the reaction B₂H₆(g) + 3O₂ (g) ⇒ B₂O₃(s) + 3H₂O(l), ΔH = -1940 kJ, we first need to calculate the number of moles of each reactant. Using the molar masses of diborane (27.67 g/mol) and oxygen (32.00 g/mol), we find:
- 25.1 g of B₂H₆ ÷ 27.67 g/mol = 0.906 moles of B₂H₆
- 57.0 g of O₂ ÷ 32.00 g/mol = 1.781 moles of O₂
Since the reaction requires three times as many moles of oxygen as diborane to react completely, oxygen is in excess, and diborane is the limiting reagent. Thus, the reaction heat is governed by the amount of diborane. By the stoichiometry of the reaction, 1 mole of B₂H₆ releases 1940 kJ. So the heat released from 0.906 moles of B₂H₆ is:
ΔH = 0.906 moles × -1940 kJ/mole = -1757.64 kJ
Approximately -1758 kJ of heat will be transferred during the reaction of 25.1 g of diborane with 57.0 g of oxygen under the given conditions.