My balanced equation is:

a) 2C8H18(g) + 25O2(g) -> 16CO2(g) + 18H2O(g)

b) 0.290 mol of octane is allowed to react with 0.670mol of oxygen. Which is the limiting reactant? (((Which I found Oxygen to be the limiting reactant.)))

I do not understant how to work out "c" & "d":

c)How many moles of water are produced in this reaction?
H2O produced: ____mol

d)After the reaction, how much octane is left?
moles of C8H18 remaining: ___mol

Could someone please help me?!

it worked out beautifully. Thank you!

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Certainly! I'd be happy to help you with parts c and d of your question. Let's break it down step by step:

c) How many moles of water are produced in this reaction?
From the balanced equation:
2C8H18(g) + 25O2(g) -> 16CO2(g) + 18H2O(g)

We can see that for every 2 moles of octane (C8H18), 18 moles of water (H2O) are produced. So the mole ratio is 18 moles of water for every 2 moles of octane.

To find the moles of water produced, you need to determine the number of moles of octane that reacted based on the given information. You stated that the limiting reactant is oxygen (O2), so you need to find the moles of octane required to react with all of the given moles of oxygen.

Given:
Moles of octane (C8H18) = 0.290 mol
Moles of oxygen (O2) = 0.670 mol

From the balanced equation, we can see that the mole ratio between C8H18 and O2 is 2:25. This means that for every 2 moles of C8H18, we need 25 moles of O2.

To find the moles of octane needed to react with all of the given moles of oxygen, set up a ratio using the mole ratio from the balanced equation:

(0.670 mol O2) * (2 mol C8H18 / 25 mol O2) = 0.0536 mol C8H18

So, 0.0536 moles of octane are needed to react with 0.670 moles of oxygen. Since you have 0.290 moles of octane, which is less than 0.0536 moles needed, it means that all of the octane will be consumed during the reaction.

Since 2 moles of octane react to produce 18 moles of water, we can calculate the moles of water produced by multiplying the moles of octane consumed (0.290 moles) by the mole ratio:

0.290 mol C8H18 * (18 mol H2O / 2 mol C8H18) = 2.610 mol H2O

Therefore, 2.610 moles of water are produced in this reaction.

d) After the reaction, how much octane is left?

Based on the information we just calculated, all of the octane is consumed during the reaction. Therefore, there is no octane left after the reaction.

So the moles of C8H18 remaining is 0 mol.

I hope this helps! Let me know if you have any further questions.

If oxygen is the limiting reagent (I didn't check that), then mols H2O produced must be 0.670 mols O2 x (18 mols H2O/25 mols O2) = ?? mols H2O.

For part d, you just have another stoichiometry problem but it might sound something like this. How much octane is needed to react with 0.670 mol oxygen in the equation written. So
mols C8H18 used = mols O2 x (2 mols C8H18/25 mols O2) = ??
Then what you started with minus ?? in the last calculation will be the amount octane left. Post your work if you get stuck.