How many moles of $O_{2}$ $O_2$ are required to generate 18 moles of $H_{2} O$ $H_2O$ in the reaction $2 C_{8} H_{18} + 25 O_{2} \to 16 C O_{2} + 18 H_{2} O$ $2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O$ ?

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Answer 1 · 2016-04-12T13:18:10

You will need 25 moles, but this is not based on the equation alone.

$18 m o l H_{2} O x (\frac{25 m o l O_{2}}{18 m o l H_{2} O}) = 25 m o l O_{2}$ $18mol H_2 O x ((25 molO_2)/ (18molH_2 O)) = 25 mol O_2$

Moles of water cancel, leaving you with moles of Oxygen.
Factor out 18/18=1.
25 x 1 = 25

How many moles of O_2O2O_2 are required to generate 18 moles of H_2OH2OH_2O in the reaction 2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O2C8H18+25O2→16CO2+18H2O2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O?