How many moles of water are produced from 0.100 mol pentane, C5H12?

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C5H12 (g) + O2 (g) —-> CO2 (g) + H209 (g) ( Needs Balance )

Please explain how to find the answer also.Thanks!

2 Answers

  • Balanced equation: C5H12(g) + 8 O2(g) -> 5 CO2(g) + 6 H2O

    Given mol C5H12 x mole ratio(mol H2O/mol C5H12) = mol H2O

    The mole ratio is obtained from the coefficients of the BALANCED equation: 6 mol H2O : 1 mol C5H12

    0.100 mol C5H12 x (6 mol H2O/1 mol C5H12) = 0.6 mol H2O

    Sig figs… 0.600 mol H2O

  • Ok, first step is to set up a balanced equation with which you can use stoicheometry to go from pentane to water.

    in this case the balanced equation looks like:

    1(C5H12)+8(O2)–>5(CO2)+6(H20)

    What this tells you is that for every one mole of pentane burned 6 moles of water is released, making the stoicheometric factor (6molH20)/(1molC5H12)

    you can then multiply .100molC5H12 by (6molH20)/(1molC5H12) , the “molC5H12” cancels out and you are left with .600 moles of water

    ——>.600 moles of water<——-

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