propane (c3h8) is burned in oxygen to produce carbon dioxide and water. The heat of combustion of propane is -2012 kJ/mole. How much heat is given off when 3.0 mol C3H8 are react with 10.0 mol O2?
Thanks, and please if possible show all steps!
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 = ?
or
2012 kJ/5 mol O2 x 10.0 mol O2 used = 2012*2 = ?
To calculate the amount of heat released when 3.0 mol C3H8 react with 10.0 mol O2, we need to use the balanced chemical equation and the molar heat of combustion of propane.
The balanced chemical equation for the complete combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, we can see that every 1 mole of C3H8 reacts with 5 moles of O2. Therefore, to use the given amount of 3.0 mol C3H8, we need to determine the amount of O2 required.
Using the molar ratio from the balanced chemical equation, we can calculate the moles of O2 needed:
(3.0 mol C3H8) × (5 mol O2 / 1 mol C3H8) = 15.0 mol O2
Since we only have 10.0 mol O2, this means that O2 is in excess. The limiting reactant is C3H8.
Now, let's calculate the amount of heat released using the molar heat of combustion of propane:
Heat of combustion of propane = -2012 kJ/mol
To calculate the heat released, we need to use the moles of C3H8. Therefore, the heat released can be calculated as follows:
(-2012 kJ/mol) × (3.0 mol C3H8) = -6036 kJ
The negative sign indicates that heat is being released during the combustion reaction.
Therefore, when 3.0 mol C3H8 reacts with 10.0 mol O2, 6036 kJ of heat is given off.