what is the chemical equation when ethanol is combusted in an excess supply of oxygen?

for this question in the solutions they have written ethanol as ch3ch2oh, if i wrote c2h6o for ethanol reactant is that fine?

    chemistry1111 Hey, im not too sure if its fine using the molecular formula in vcaa exam, maybe someone can confirm for you. But i personally would not recommend it as a ketone or aldehyde could also have the same molecular formula C2H6O, so i think it is just better to use the semi struc version since its clearer what the molecule is.

      chemistry1111 I think VCAA accepted C2H6O before, but I wouldn't recommend it cuz the molecular formula could be that of an isomer with different functional groups. For fuel definition, I don't think VCAA will ask it, but make sure to allude to the fact that a fuel is a substance with potential energy stored within it (chemical or nuclear) that can be converted to useful forms of energy

          for question 2 of the 2022 vce exam, why isnt the answer B cause i think solutions i saw said A. could someone explain

            chemistry1111 that’s cuz you’re not burning water, energy is released by the fuel to heat its surroundings (in this case, H2O)

              Does anyone know how to do this

              Propane burns completely in oxygen according to the equation:
              C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
              The amount of energy released per tonne of
              carbon dioxide produced is determined to be
              1.68 × 104 MJ t–1.
              Determine the ΔH value, in kJ mol–1, for the thermochemical equation. (1 tonne = 106 g)

                a month later

                snowflake
                Combusting 1 tonne of C3H8 releases 1.68 × 104 MJ

                So the question is, how much energy does combusting 1 mol of C3H8 release?

                So first you need to convert 1 tonne into mol:
                1 tonne = 1 000 000 grams
                n = m/M (n = number of moles, m = mass, M = molar mass)
                M (C3H8) = 3 x 12 + 8 x 1 = 44 g/mol
                n = 1 000 000 / 44 = 22727 mol

                So if combusting 22727 mol of C3H8 releases 1.68 × 104 MJ, combusting 1 mol would release (1.68 x 104)/22727 MJ = 0.7392 MJ = 0.7392 x 1000 kJ = 739.2 kJ

                Therefore, the ΔH value, in kJ mol–1, for the thermochemical equation is 739 kJ/mol.

                  9 days later

                  is there a difference between petrodiesel and petroleum gas or are they the same thing?

                    mia46789

                    Hi, yes there is a difference. Petrodiesel is 25% aromatic hydrocarbons and 75% alkanes which are in the C10H22-C15H32 range. Petroleum gas, on the other hand, is composed of much shorter alkanes, mainly propane (C3H8) and butane (C4H10).

                    Another difference is that petrodiesel is a liquid, whereas petroleum gas, as the name suggests, is a gas, though it is usually stored under high pressures making it liquid petroleum gas (LPG).

                    Both are usually obtained from crude oil.

                        GreenAcorn I agree with the above, it's usually safe to assume that petrodiesel is dodecane (C12H26) and petroleum gas is propane. Likewise, you can assume petrol is octane

                          13 days later

                          im trying to calculate the heat of combustion for 3-methylbutan-1-ol in kj/g and kj/mol for this calorimetry experiment
                          this is the information I have:
                          mass of water (inside the metal can) (g) = 99.31 g
                          initial mass of spirit burner (3-methylbutan-1-ol) (g) = 194.16 g
                          final mass of spirit burner (g) = 193.8 g
                          initial temperature of water (celsius) = 21.3 C
                          final temperature of water (celsius) = 52.8 C
                          I've been using the textbook to try and figure it out but I'm pretty sure the answer i keep getting is wrong

                            bugsme_
                            m(alcohol used) = 194.16 g - 193.8 g = 0.36 g

                            energy released (assuming no heat loss to surroundings) = 99.31 g x 4.18 J/g/K x (52.8˚C - 21.3˚C) = 13076.1477 J

                            Heat of combustion per g = 13.076... kJ / 0.36 g = 36.3 kJ/g

                            Heat of combustion per mol = 13.076 kJ / (0.36 g / 88 g/mol) = 3196 kJ/mol

                            However, since the final answer should be rounded to 3 sig figs (as temp was to 3 sig figs) and 3200 kJ/mol would be 4 sig figs according to VCAA, you would express this as 3.20 x 103 kJ/mol

                              11 days later

                              Hello!

                              Does anyone have Chemistry AOS 1 practice sacs? I Have my sac on Thursday and my teacher has only given 2 TSSM sac's!!!

                              2 months later

                              hi,
                              we just started the topic of fuel cells and i'm very confused. what's the connection between electrolytic, primary, secondary, voltaic, electrochemical cells?? we just did reduction and oxidation and that's fine but i can't seem to understand the whole 'batteries' thing.
                              someone pls help a stressed student out

                                juliarobertsxxx Voltaic cells are electrochemical cells that convert chemical energy to electrical energy. Most of the time, VCAA wants to see the term “galvanic cell” since that’s the more common name for it. Primary and secondary cells are types of galvanic cells (primary = non rechargeable, secondary = rechargeable), whereas fuel cells are special galvanic cells that only work while the fuel is being supplied.

                                Electrolytic cells are electrochemical cells that use electrical energy to produce a chemical species through a non spontaneous reaction. When a secondary cell is being recharged, it’s an electrolytic cell

                                  3 months later

                                  Hi, this is related to subject selections but I wanted to know if it is it possible to take VCE Chem without prior exposure to it?
                                  I didn't take subjects like Environmental Science or Investigations but I was hoping to take Chem for Yr 11 but I'm worried that I'm already behind considering it is a very hard subject.
                                  I took only took Proj and Forensics this year so my Chem understanding is very minimal.

                                    c:_melly If you have an interest in Chemistry and the motivation to apply yourself, I would say go for it! That's what Year 11 Chemistry is for, familiarising you with the basic concepts before going into Year 12.