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.
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?!
4 answers
it worked out beautifully. Thank you!
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