Question

What volume in L of O2 at STP is needed to burn 1.00 L (702g) of octane

Answer 1

84733 Chemistry, Other

What volume in L of $\mathrm{O}_2$ at STP is needed to burn 1.00 L (702 g) of octane?

Answer:

$2\mathrm{C}_{8}\mathrm{H}_{18}(\mathrm{g}) + 25\mathrm{O}_{2}(\mathrm{g})\rightarrow 16\mathrm{CO}_{2}(\mathrm{g}) + 18\mathrm{H}_{2}\mathrm{O}(\mathrm{g})$

PV=nRT

n $(O_2)$ = 702g $C_8H_{18}$ x (1 mol $C_8H_{18}$ / 114.0 g $C_8H_{18}$ ) x (25 mol $O_2$ / 2 mol $C_8H_{18}$ ) = 76.97 mol

V = nRT / P

V $(\mathrm{O}_2)$ = (76.97 mol x 0.0821 Latm/molK x 291K) / 0.975 atm = 1886 L

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