Pentaborane-9(B5H9) IS a colorless highly reactive liquid that will burst into flames when exposed to oxygen.The reaction is '
2B5H9(l)+ 12O2--> 5B2O2(s)+9H2O(l)
Calculate the kilojoules of heat released per gram of the compound reacted with oxygen.The standard enthalpy of formation B5H9 is 73.2kJ/mol
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To calculate the kilojoules of heat released per gram of Pentaborane-9 (B5H9) reacted with oxygen, we need to follow these steps:
Step 1: Calculate the molar mass of B5H9.
Boron (B) has an atomic mass of approximately 10.81 g/mol, and Hydrogen (H) has an atomic mass of approximately 1.01 g/mol. Since there are 5 boron atoms and 9 hydrogen atoms in each molecule of B5H9, we can calculate the molar mass as follows:
Molar mass of B5H9 = (5 * molar mass of B) + (9 * molar mass of H)
Molar mass of B5H9 = (5 * 10.81 g/mol) + (9 * 1.01 g/mol)
Molar mass of B5H9 = 54.05 g/mol + 9.09 g/mol
Molar mass of B5H9 = 63.14 g/mol
Step 2: Calculate the moles of B5H9 reacted.
We need to convert the mass of B5H9 reacted to moles. Since we don't have the mass given in the question, we'll assume it's 1 gram for this example.
Moles of B5H9 = Mass of B5H9 / Molar mass of B5H9
Moles of B5H9 = 1 g / 63.14 g/mol
Moles of B5H9 = 0.0158 mol
Step 3: Calculate the heat released per mole of B5H9 reacted.
Using the balanced equation provided in the question, we can see that for every 2 moles of B5H9 reacted, 73.2 kJ of heat is released.
Heat released per mole of B5H9 = 73.2 kJ / 2 mol
Heat released per mole of B5H9 = 36.6 kJ/mol
Step 4: Calculate the heat released per gram of B5H9 reacted.
To find the heat released per gram, we need to divide the heat released per mole by the moles of B5H9 reacted.
Heat released per gram of B5H9 = Heat released per mole / Moles of B5H9
Heat released per gram of B5H9 = 36.6 kJ/mol / 0.0158 mol
Heat released per gram of B5H9 ≈ 2317.09 kJ/g
Therefore, approximately 2317.09 kilojoules of heat are released per gram of Pentaborane-9 (B5H9) reacted with oxygen.