The combustion of octane...
Chemistry

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The complete combustion of octane, C8H18, the main component of gasoline, proceeds as shown below.
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)
(a) How many moles of O2 are needed to burn 1.00 mol of C8H18?
(b) How many grams of O2 are needed to burn 1.75 g of C8H18?
(c) Octane has a density of 0.692 g/mL at 20°C. How many grams of O2 are required to burn 3.35 gal of C8H18?
Thank you for the opportunity to help you with your question!
a) 25 moles of O2 needed to burn 2 moles of C8H18. Hence 12.5 mole of O2 needed to burn 1 mole.
b)2.626g of O2
c)8.476g of O2
Could you explain your steps for b and c please?
Ok,
12.5 moles of O2 needed to burn 1 mole of C8H18.I mole of Octane is 114.23 g
so 1.75g is:
1.75/114.23=0.015moles of Octane.
Using the previous relationship. If 1 mole of Octane is burnt using 12.5 moles of O2, then 0.015moles would be burnt using : 0.015*12.5= 0.1915 moles of O2 needed.
1 mole of O2 is 16g
0.1915moles is: 0.1915*16=3.064g of Oxygen.
I must have made a mistake somewhere because of the rush. Thanks for asking for clarification. Let me explain c now.
There are 0.692g/ml . And if 0.692g= 2619.5ml, and 1ml=1g, hence 2619.5ml=2619.5g
If 114.23g=1mole of Octane, 2619.5 would be 22.93 moles.
If Imole of Octane needed 12.5 moles of O2 to burn, then 22.93 moles will need 286.65moles to burn.
IF 1mole of O2 is 16g, then 286.65 moles would be 4586.36g
Hence we need 4586.36g.
I was in a rush before because of the timing. I hope it is clear now. I am always happy to help.
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