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owlet

  • one year ago

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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  1. cuanchi
    • one year ago
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    That equation is not balanced

  2. cuanchi
    • one year ago
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    \(2C_8H_{18(g)}+25O_{2(g)}→16CO_{2(g)}+18H_2O_{(g)}\)

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

  4. cuanchi
    • one year ago
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    yes

  5. cuanchi
    • one year ago
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    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

  6. owlet
    • one year ago
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    nvm, i know which part i did wrong.. i used 0.63 moles of oxygen instead of 0.86 thanks anyway :)

  7. cuanchi
    • one year ago
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    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

  8. owlet
    • one year ago
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    i honestly don't know where i get that 0.63, i must be too tired now xD

  9. cuanchi
    • one year ago
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    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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