Hey Guys these 2 questions are really bugging me can someone EXPLAIN them to me
1)
The molar enthalpy of combustion of camphor, C10H16O, is -5898 kJ/mol. How much energy is released on combustion of 1 gram of camphor?
2) Determine the enthalpy of reaction when 1 mole of ethene reacts with oxygen to form carbon dioxide and water. Using the bond energies (attached)
Thankyou guys !!!
Hey! Here's how we work out the first one. We know that the heat of combustion is

But, we want the answer to be in kJ/g, right? To get from kJ/mol to kJ/g, we would need to do something like this

Cool! So, what do we know that has units of mol/g (or, rather, g/mol and then we can just divide it)?. The molar mass! The molar mass of Camphor is

So,

A more intuitive way of doing this is thinking about what kJ per mol actually means. It means kJ
per 152.23 grams! So, just divide by the grams to get kJ per gram

Now, onto your second, question, which I really don't know about
WAY OUTSIDE OF HSCIt looks like we do it like this. Our equation is
}+3O_{2(g)} \rightarrow 2CO_{2(g)}+2H_2O_{(l)})
Okay. Now it gets tough. In Ethene, there is one double-bonded Carbon-Carbon bond, and four single-bonded Carbon-Hydrogen bond. From the formula sheet, the total bond energy will be

As there is only one mole, the bond energy of Ethene is 2276 kJ. In Oxygen, there is one double-bonded Oxygen-Oxygen bond.

As there are three moles, the bond energy will be 1497kJ.
Now, we need to look at the products. Carbon dioxide has two double-bonded Carbon-Oxygen bonds. From the formula sheet

Since there are two moles, the bond energy will be 3196 kJ. In Water, there are two single-bonded Hydrogen-Oxygen bonds. Thus,

As there are two moles of water, the bond energy will be 1840 kJ.
Adding up all the reactant energies, we get 3773 kJ. Adding up all the product energies, we get 5036 kJ. To find the enthalpy change, we
subtract the products from the reactants, giving us

Note that I've never done this before, so I may be very wrong.