-14 J/Mol
The heat of reaction per mole can be calculated by dividing the energy produced by the number of moles. In this case, 84 J of energy produced by 6 moles of the compound gives a heat of reaction of 14 J/mol.
84 J/6 moles = 14 J/mole = ∆H
The enthalpy change (ΔH) per mole can be found by dividing the energy produced by the moles of the compound. In this case, ΔH = 84 J / 6 mol = 14 J/mol. Therefore, the enthalpy change per mole of the compound is 14 J/mol.
The enthalpy change for the reaction would be -8 J/mol, as it is the energy change per mole of the compound reacted.
This is not a common reaction at standard temperature and pressure.
The heat of reaction per mole can be calculated by dividing the energy produced by the number of moles. In this case, 84 J of energy produced by 6 moles of the compound gives a heat of reaction of 14 J/mol.
84 J/6 moles = 14 J/mole = ∆H
12 J/3 moles = 4 J/mole. Thus, H of reaction in kJ/mole = 0.004 kJ/mole
-14 J/mol
The enthalpy change (ΔH) per mole can be found by dividing the energy produced by the moles of the compound. In this case, ΔH = 84 J / 6 mol = 14 J/mol. Therefore, the enthalpy change per mole of the compound is 14 J/mol.
The enthalpy change for the reaction would be -8 J/mol, as it is the energy change per mole of the compound reacted.
Since acetylene (C2H2) has a stoichiometry of 2 moles of acetylene to produce 2 moles of CO2, three moles of acetylene would produce 3 moles of CO2. The reaction with excess oxygen ensures that all the acetylene is fully converted to CO2.
This is not a common reaction at standard temperature and pressure.
N2 + 3H2 -----> 2NH3 so 3 moles of hydrogen produce 2 moles of ammonia. Therefore 12.0 moles of hydrogen will produce 8 moles of ammonia.
If 5.0 moles of NH3 are produced 2.5 moles of N2 are used.
Carbon dioxide is the limiting reagent.
For the reaction 2 H2 + O2 -> 2 H2O, we see that 1 mole of O2 produces 2 moles of H2O. Therefore, to produce 10.2 moles of H2O, we would need 5.1 moles of O2.