Question

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.

Answer 1

That equation is not balanced

Answer 2

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

Answer 3

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

Answer 4

yes

Answer 5

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

Answer 6

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

thanks anyway :)

Answer 7

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

Answer 8

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

Answer 9

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