Solve for moles grams

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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?

Jul 24th, 2015

Thank you for the opportunity to help you with your question!

(a) 25 moles of O2 are needed to burn 2 moles of C3H18. It is noted from the given reaction.

Hence, 12.5 moles of O2 will be needed to burn 1 mole of C3H18.

(b) 2C8H18 = 2(12*8+18*1) = 228

25O2 = 25*16*2 = 800

Hence, O2 required to burn 1.75 gm of C8H18 = (800/228)*1.75 = 6.14gms

(c) Weight of 3.35 gallon of octane = 3.35*1000*0.692ml*gm/ml = 2318.2gm

Hence O2 needed = (800/228)*2318.2 = 8134.04gms


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Jul 24th, 2015

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