Answer to Question #138202 in General Chemistry for Stacey

Question

The combustion of octane, C8H18,
proceeds according to the reaction shown.

2C8H18(l)+25O2(g)⟶16CO2(g)+18H2O(l)


Calculate the volume of carbon dioxide produced when 466 moles
of octane combusts at 29.0 °C
and 1.00 bar.

Answer 1

First of all let's calculate the volume of carbon dioxide produced at STP

stochiometric coefficient ratio of octane and carbon dioxide is 2:16=1:8

466 moles of octane will give 8$\times$466 moles of carbon dioxide

and one mole is equivalent to 22.4 liters at STP

hence total volume at STP = $8\times466\times22.4=83507.2\ liters$

Since STP refers to 0$^o$ C and we have to calculate volume at 29^o C

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$\frac{83507.2}{273}=\frac{V_2}{302}$

$V_2=92377.92liters$