Answer:
-1670.24 kJ
QUICK Explanation:
ΔH * moles of limiting reactant = Energy produced
-572 kJ/mol * 2.92 moles of O2 = -1670.24 kJ
LONGER EXPLANATION :
2 H2 + O2 -> 2 H2O
ΔH * moles of limiting reactant = Energy produced
enthalpy change or heat of reaction formula
1. Calculate the number of moles of oxygen (O2):
Number of moles = mass / molar mass
Number of moles of O2 = 93.5 g / 32.00 g/mol
≈ 2.92 mol O2
2. Calculate the number of moles of hydrogen (H2):
Number of moles = mass / molar mass
Number of moles of H2 = 13.2 g / 2.02 g/mol
≈ 6.53 mol H2
3. Determine the limiting reactant:
According to the balanced equation,
2 moles of H2 react with 1 mole of O2.
Calculate the moles of O2 based on the moles of H2:
(6.53 mol H2) / (2 mol H2/O2) = 3.27 mol O2
we need 3.27 mol O2 to react with the available H2
BUT only have 2.92 mol of O2 available
O2 is the limiting reactant
4.
Calculate the heat given off by assuming the complete consumption of the limiting reagent
calculate the amount of energy produced using the given enthalpy change (ΔH):
Energy produced = ΔH * moles of limiting reactant
Energy produced = ΔH * moles of O2 reacted
Calculate the energy produced using the given enthalpy change (ΔH):
Energy produced = ΔH * moles of O2 reacted
= -572 kJ/mol * 2.92 mol
≈ -1670.24 kJ
Therefore, approximately -1670.24 kJ of energy is produced when 93.5 grams of oxygen react with 13.2 grams of hydrogen in the given reaction. Note that the negative sign indicates that the reaction is exothermic (energy is released).
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