First you must determine which is the limiting reagent.
C3H8 + 5O2 ==> 3CO2 + 4H2O
3 mol C3H8 x (3 mol CO2/1 mol C3H8) = 3*3/1 = 9 mol CO2
10 mol O2 x (3 mol CO2/5 mol O2) = 10*3/5 = 6 mol CO2
So the limiting reagent is O2. You are reacting 5 mol O2 to produce 6 mols CO2.
You get 2012 kJ/5*32 g oxygen. Therefore, for 320 g (10*32 g O2) you will get
(2012 kJ/160) x 320 = ?
You could do it another way with C3H8 but you must determine how much of the C3H8 reacted since some of it is left. 10 mols O2 x (1 mol C3H8/5 mol O2) = 2 mol C3H8 reacted.
Then (2012 kJ/44 g C3H8) x (2*44 g C3H8) = ?
Hope this helps. Next time tell me what your trouble is so I can concentrate on that. This way I'm just doing your work for you and you're trying to figure out where you went wrong.
ok will do, I just honestly had no idea how to do this. Can you explain how you got "2012 kJ/5*32 g oxygen. Therefore, for 320 g (10*32 g O2) you will get
(2012 kJ/160) x 320 = ? " this part?
thanks :)I don't understand where the grams came from, like I know it is from O2, but why can you not use moles? why do you have to use grams?
You don't have to use grams and you CAN use mols. I don't and the reason I don't use mols is it keeps me from making a mistake.
The equation tells you that you get 2012 kJ/5 mols O2 or 2012 kJ/1 mol C3H8. (and this is where I make my mistake---I am want to reason that since it is 2012 kJ/MOL that must mean 2012 kJ/mol C3H8 OR 2012 kJ/mol O2 and it isn't. It's 2012 kJ/1 mol C3H8 or 2012 kJ/5 mol O2. By converting to grams I don't make the error).
5 mols O2 is 5*32 = 160 g O2.
1 mol C3H8 is 44 g.
So you get 2012 kJ/160 g O2 or you get 2012 kJ/44 g C3H8. Same statement.
You want to know, for 10 mols O2, how much energy is released so that is
(2012 kJ/160g O2) x 320 g O2 = ?
It may be a little less confusing to you if we don't convert to grams (the problem is in mols anyway)
So you used 10.0 mols O2 and 2 mols C3H8.
2012 kJ/1 mol C3H8 x 2 mol C3H8 used = 2012*2 = ?
2012 kJ/5 mol O2 x 10.0 mol O2 used = 2012*2 = ?
Chemistry - A small bottle of propane gas (C3H8) contains 400.0 grams of propane...
AP chemistry - Propane (C3H8) is burned in oxygen to pro- duce carbon dioxide ...
chemistry - How many grams of propane (C3H8) must be burned to release 3600kJ of...
Chemistry Help!! - Propane(C3H8), the major component of LP game, is often used ...
Chemistry - Consider the combustion of propane: C3H8(g)+5O2(g) yields 3CO2(g) + ...
chemistry - The combustion of propane is given by the following reaction: C3H8 (...
chemistry question - What mass of propane (C3H8) is needed to produce 346 g ...
Chemistry - Consider the combustion of propane: C3H8 + 5CO2 -> 3CO2 + 2H20. ...
Chemistry - Oops I posted the right question but wrong answer options... What ...
Science - Consider the combustion of propane: C3H8(g)+5O2(g) yields 3CO2(g) + ...