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- owlet

Determine the balanced chemical equation for this reaction.
\(2C_8H_{18(g)}+25O_{2(g)}→16CO_{2(g)}+18H_2O_{(g)}\)
0.360 mol of octane is allowed to react with 0.860 mol of oxygen. Which is the limiting reactant? My ans is oxygen because I only got 25.6 g while octane has 41.04 g.
How many moles of water are produced in this reaction?
I need help with this one... here is my solution:
0.63 mol O2 x (18 mol H2O /25 mol O2) = 0.576 mol of H2O
but it is wrong.

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- owlet

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- Cuanchi

That equation is not balanced

- Cuanchi

\(2C_8H_{18(g)}+25O_{2(g)}→16CO_{2(g)}+18H_2O_{(g)}\)

- owlet

oops! I missed the coefficients. but I have the same balanced equation on my paper.
Do you know how to work this one out?

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- Cuanchi

yes

- Cuanchi

the oxygen is the limiting reactant
0.360 mol C8H18 x 16/2=2.88 moles ofCO2
0.860 moles O2 x 16/25 = 0.55 moles CO2

- owlet

nvm, i know which part i did wrong.. i used 0.63 moles of oxygen instead of 0.86
thanks anyway :)

- Cuanchi

there is 0.860 moles of O2!!! I dont know why you put 0.63
0.860 mol O2 x (18 mol H2O /25 mol O2) = 0.6192 mol of H2O

- owlet

i honestly don't know where i get that 0.63, i must be too tired now xD

- Cuanchi

yea happens to me I am sometimes number dyslexic and I mix the 6, 8 and 9 , I have to double check

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