2 B5H9(l)+12 O2(g)---->5 B2O3(s)+9 H2O(l)
Calculate the kilojoules of heat released per gram (kJ/g) of the compound reacted with oxygen. The standard enthalpy of formations of B5H9(l), B2O3(s), and H2O(l) are 42.84, -1271.94, and -285.83 kJ/mol, respectively.
delta Hrxn = (n*sum delta H products)-(n*sum delta H reactants)
To calculate the kilojoules of heat released per gram (kJ/g) of the compound reacted with oxygen, we need to determine the number of moles of each compound and then calculate the heat released.
First, let's determine the molecular weights of B5H9, B2O3, and H2O:
- Molecular weight of B5H9 = (5*10.81) + (9*1.00784) = 55.11 g/mol
- Molecular weight of B2O3 = (2*10.81) + (3*16.00) = 69.62 g/mol
- Molecular weight of H2O = (2*1.00784) + (16.00) = 18.02 g/mol
Next, let's calculate the number of moles of each compound:
- Moles of B5H9 = mass (g) / molecular weight (g/mol)
- Moles of B5H9 = 1 g / 55.11 g/mol = 0.018 moles
Using the balanced chemical equation, we can determine the moles of B2O3 and H2O produced:
- Moles of B2O3 = 5/2 * Moles of B5H9
- Moles of H2O = 9/2 * Moles of B5H9
Now, let's calculate the heat released:
- Heat released = Moles of B2O3 * ∆Hf(B2O3) + Moles of H2O * ∆Hf(H2O)
∆Hf(B2O3) = -1271.94 kJ/mol
∆Hf(H2O) = -285.83 kJ/mol
- Heat released = (5/2 * 0.018) * (-1271.94 kJ/mol) + (9/2 * 0.018) * (-285.83 kJ/mol)
Calculating the heat released:
- Heat released = (0.045 * -1271.94 kJ) + (0.081 * -285.83 kJ)
- Heat released = -57.2393 kJ + -23.1543 kJ
- Heat released = -80.3936 kJ
Finally, let's calculate the kilojoules of heat released per gram:
- kJ/g = Heat released / mass (g)
- kJ/g = -80.3936 kJ / 1 g
Therefore, the kilojoules of heat released per gram (kJ/g) of the compound reacted with oxygen is -80.3936 kJ/g.
To calculate the kilojoules of heat released per gram (kJ/g) of the compound reacted with oxygen, we need to first calculate the moles of each compound involved in the reaction. Then, we can use the stoichiometry of the reaction and the standard enthalpy of formation values to calculate the heat released. Here's how you can do it step by step:
1. Calculate the number of moles of each compound:
- B5H9(l): Given that the molar mass of B5H9 is 54.27 g/mol, the number of moles of B5H9 is:
moles of B5H9 = mass of B5H9 (g) / molar mass of B5H9
- O2(g): Since O2 is a diatomic molecule, the molar mass of O2 is 32 g/mol. The number of moles of O2 can be calculated using the ideal gas law or by dividing the amount of oxygen given (in grams) by the molar mass.
- B2O3(s): Given that the molar mass of B2O3 is 69.62 g/mol, the number of moles of B2O3 is:
moles of B2O3 = mass of B2O3 (g) / molar mass of B2O3
- H2O(l): Given that the molar mass of H2O is 18.01 g/mol, the number of moles of H2O is:
moles of H2O = mass of H2O (g) / molar mass of H2O
2. Determine the limiting reactant:
Determine the limiting reactant by comparing the moles of B5H9 and O2. The reactant that produces fewer moles of product is the limiting reactant.
3. Calculate the moles of product formed:
- Using the stoichiometry of the balanced equation, calculate the moles of B2O3 and H2O formed.
moles of B2O3 = (moles of limiting reactant) * (5 mol B2O3 / 2 mol B5H9)
moles of H2O = (moles of limiting reactant) * (9 mol H2O / 2 mol B5H9)
4. Calculate the heat released:
- Use the given standard enthalpy of formation values to calculate the heat released.
heat released = (moles of B5H9) * (ΔHf of B5H9) + (moles of B2O3) * (ΔHf of B2O3) + (moles of H2O) * (ΔHf of H2O)
5. Finally, calculate the kilojoules of heat released per gram (kJ/g) of the compound reacted with oxygen:
kJ/g = heat released (kJ) / mass of the B5H9 (g)
Just plug in the values and perform the calculations to find the answer.