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Stoichiometry is a fundamental concept in chemistry that relates to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the balance and proportion of elements involved in the process. In the given chemical reaction of diborane combusting in oxygen, stoichiometry helps us understand how the moles of reactants convert into moles of products. Using the balanced equation,

• 1 mole of \( \text{B}_2\text{H}_6 \) reacts with 3 moles of \( \text{O}_2 \)
• Results in 1 mole of \( \text{B}_2\text{O}_3 \) and 3 moles of \( \text{H}_2\text{O} \)

These relationships indicate that for every mole of diborane combusted, an equal number of moles of \( \text{B}_2\text{O}_3 \) and three times that of water are formed. This mole ratio is crucial when calculating the expected outcomes of the reaction, allowing precise calculations of amounts generated or used.

Molar mass is an essential concept that tells us the mass of one mole of a substance, allowing for the conversion between grams and moles. It is determined by summing the atomic masses of all atoms in a molecular formula. For diborane, \( \text{B}_2\text{H}_6 \),

• The atomic mass of Boron (B) is approximately 10.81 amu
• The atomic mass of Hydrogen (H) is approximately 1.01 amu

Using this, the molar mass of diborane is calculated as:\[(2 \times 10.81) + (6 \times 1.01) = 27.62 \text{ g/mol}\] This value lets us convert the mass of diborane provided in the exercise into moles:\[\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}}\]Thus, for 54.0 g of diborane, the moles are:\[\frac{54.0 \text{ g}}{27.62 \text{ g/mol}}\] Providing the precise number of moles needed in the stoichiometric calculations.

The concept of heat released refers to the amount of energy liberated when a reaction occurs, commonly expressed in kilojoules (kJ). Enthalpy change (\( \Delta H \)) signifies this energy transfer, and it is negative for exothermic reactions where heat is released. In the case of the combustion of diborane, the enthalpy change is given as\[\Delta H = -2035 \text{ kJ/mol}\] This means that for every mole of diborane combusted, 2035 kJ of heat is released. By knowing the number of moles of diborane,

• Heat released = \( \Delta H \times \text{moles of } \text{B}_2\text{H}_6 \)
• For 1.954 moles calculated, the heat released is \(-2035 \text{ kJ/mol} \times 1.954 \text{ mol} \)

This results in a total of -3973 kJ, indicating a substantial release of energy during the combustion process, with the negative sign emphasizing the exothermic nature of the reaction.