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Let's assume all propanol reacts fully with the oxygen.

First let's set up the reaction equation.

2 C3H8O1 + 9 O2 --> 8 H2O + 6 CO2

2 mole propanol reacts with oxygen to form 6 mole of CO2.

Thus the mole ratio is 2:6 = 1:3

So 0.136 mole of propanol forms 0.136 * 3 = 0.408 mole CO2

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8y ago
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8y ago

The excess oxygen ensures that full combustion to dioxide occurs and no carbon monoxide nor carbon is produced. The reaction is given by:

2 C3 H7 OH + 9 O2 → 6 C O2 + 8 H2 O

(Spaces used to make the formulae easier to read as subscripts not available for the numbers)

So for every 3 moles of propanol 6 moles of carbon dioxide are produced; thus there will be:

0.136 ÷ 2 × 6 = 0.136 × 3 = 0.408 moles of carbon dioxide produced.

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13y ago

0.95 - 0.954

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14y ago

0.680

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14y ago

58.5 - 58.9

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Q: When 0.136 moles of 1-pentanol react with excess oxygen in a combustion reaction?
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