6.3 g of pentaborane, B5H11 is burnt in the presence of O2 to yield H2O(l) and boron oxides. The reaction was performed in an open container and 157.2 kJ of heat was produced at standard conditions. What is the combH of pentaborane in kJ/mol to 2 d.p?
Question
6.3 g of pentaborane, B5H11 is burnt in the presence of O2 to yield H2O(l) and boron oxides. The reaction was performed in an open container and 157.2 kJ of heat was produced at standard conditions. What is the combH of pentaborane in kJ/mol to 2 d.p?
Solution
To find the combustion enthalpy (combH) of pentaborane, we first need to know the molar mass of pentaborane (B5H11).
The molar mass of boron (B) is approximately 10.81 g/mol and that of hydrogen (H) is approximately 1.01 g/mol. Therefore, the molar mass of pentaborane is (510.81 g/mol) + (111.01 g/mol) = 54.05 g/mol + 11.11 g/mol = 65.16 g/mol.
Next, we need to convert the amount of pentaborane used in the reaction from grams to moles. This is done by dividing the mass of pentaborane by its molar mass:
6.3 g / 65.16 g/mol = 0.0967 mol
The heat produced in the reaction is given as -157.2 kJ. The negative sign indicates that the reaction is exothermic (i.e., it releases heat).
The combustion enthalpy (combH) of pentaborane is then calculated by dividing the heat produced by the amount of pentaborane used in moles:
-157.2 kJ / 0.0967 mol = -1625.54 kJ/mol
Therefore, the combustion enthalpy (combH) of pentaborane is -1625.54 kJ/mol to 2 decimal places.
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