When 1 mole of ethylene (C2H4) is burned at constant pressure, 1410 kJ of energy is released as heat. Calculate (delta) H for a process in which 10.0 of ethylene is burned at constant pressure

C2H4 + 3O2 ==> 2CO2 + 2H2O

10.0 what? Change 10? to mols if it isn't already in mols.
dH = 1410 kJ/mol x (# mols) = ?

To calculate ΔH (change in enthalpy) for a process, you need to use the concept of stoichiometry. Here's how you can calculate it for the given question:

1. Start by writing the balanced chemical equation for the combustion of ethylene (C2H4):
C2H4 + 3O2 -> 2CO2 + 2H2O

2. Determine the molar enthalpy change (ΔH) for the combustion of ethylene. According to the balanced equation, 1 mole of ethylene (C2H4) produces 1410 kJ of energy as heat. Therefore, the ΔH for 1 mole of ethylene is -1410 kJ.

3. Calculate the ΔH for the given process, which involves burning 10.0 moles of ethylene. Since the ΔH is proportional to the number of moles, you can multiply the ΔH for 1 mole of ethylene by the number of moles involved in the process:
ΔH = (-1410 kJ/mol) * (10.0 mol) = -14100 kJ

Thus, the ΔH for the process in which 10.0 moles of ethylene are burned at constant pressure is -14100 kJ. The negative sign indicates that energy is released as heat in the process.