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) ratio=2:25

25/2 =12.5moles

b) 12x8 + 18=114

32x1.75/114

=0.49122g

c) m=density x volume

=0.692x3.35

=2.3182g

Could you explain ur answer I dont get how you arrived

first of all you get the ratio of octane to oxygen in the reaction, which is 2:25

using the same ratio, we get the no. of moles of O2 by (1molex25)/2=12.5moles

b) here, we use molar masses, O2=32g/mol , C8H18=12x8+1x18=114g/mol

since carbon=12 and hydrogen=1

...You need to use more precise numbers like 12.01 and 1.01 please

those ones are very precise, 12 and 1 are the rounded off ones, though still leads to almost the same answer value, but instead of 12 and 1 you may insert 12.01 and 1.01 respectively

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