How many moles of oxygen are needed for the complete combustion of 0.750 moles of propane?

Firstly, write a balanced equation with states:


A few things: how I knew that carbon dioxide and water were the products was because of the key phrase complete combustion. You simply have to know these types of general reactions (if you don't know already).

How I balanced the equation was via a CHOS methods.
C stands for carbon – balance the atoms first.
H stands for hydrogen – balance the atoms second (usually)
O stands for hydrogen – balance the atoms third (usually)
S stands for states – you just have to know them for the reagents and products or try to use your knowledge of bonding, boiling/melting points, etc. to deduce the states.
Sometimes you may come up with a fraction/decimal in the coefficient of a reagent/product. I know that some teachers/people are fine with that, but at least for me, I make sure all coefficients are whole numbers – so say a coefficient is , you would multiply your whole equation by 2 to counteract this.

Next, we use stoichiometry, which essentially concerns the chemical relationship between amounts (mols)/masses, etc. of substances within a reaction.

If 0.750 mol of propane reacts, we look at the coefficients of our reagents and see that oxygen gas has a 5 in front of it. THis means that the amount of moles required for the reaction to occur is 5x that of propane.

So 0.750 x 5 = 3.75 mol of oxygen gas is needed for the complete combustion of 0.750 moles of propane.

Hopefully this is correct! :P Or if not, hopefully the process of working helped you!

Sorry i dont understand the question can you rephase it?