Calc ∆Hf (kJ/mol) for C3H8 + 5 O2 -> 3 CO2 + 4 H2O

[chrisisone]

Determine the total enthalpy of the reactants and the total enthalpy of the products. Record these values in Table C of the Student Worksheet.

I need to know the total number of products and reactants in ∆Hf (kJ/mol) The substances are:
C3H8+ 5 O2= 3 CO2 + 4 H2O please help me>

please pm me or relpy wth thread please I need hep

 

[jvicens]

Based on the chemical equation provided, there are 2 reactants (C3H8 and O2) and 2 products (CO2 and H2O). To determine the total enthalpy of the reactants and products, you will need to use the standard enthalpy of formation (∆Hf) values for each substance. These values can be found in reference tables or online databases.

To calculate the total enthalpy of the reactants, you will need to multiply the number of moles of each substance by its respective ∆Hf value and then add them together. For example, for C3H8, the ∆Hf value is -103.8 kJ/mol, so the total enthalpy of the reactants would be: (1 mol C3H8)(-103.8 kJ/mol) + (5 mol O2)(0 kJ/mol) = -103.8 kJ.

Similarly, to calculate the total enthalpy of the products, you will need to multiply the number of moles of each substance by its respective ∆Hf value and then add them together. For CO2, the ∆Hf value is -393.5 kJ/mol, and for H2O, the ∆Hf value is -241.8 kJ/mol. So the total enthalpy of the products would be: (3 mol CO2)(-393.5 kJ/mol) + (4 mol H2O)(-241.8 kJ/mol) = -2028.7 kJ.

Remember to record these values in Table C of the Student Worksheet. Hope this helps!