calculating volume of co2

Question

what is the volume of carbon dioxide gas formed at s.t.p when 2.15 g of liquid hexane is completely burnt in the air. (H=1, C=12, 1 mole of gas occupies a volume of 22.4dm3 at s.t.p)

Answer 1

22.4 dm³=22.4 L, the molar volume of a gas at STP.. Combustion equation is C₆H14 + O₂--->6CO₂ +7H₂O, 2C₆H₁₄ + 19O₂--->12CO₂ + 14H₂O. 2.15 grams hexane=2.15 grams/86 grams/mole=0.025 moles hexane.

0.025 moles hexane X 12 moles CO₂/2 moles C₆H14=0.15 moles CO_2. At STP, 1 mole gas occupies 22.4L (22.4 dm³). so 0.15 moles X 22.4 dm³/mole=3.36 dm³.

Answer 2

First, we must write the correctly balanced equation for the reaction taking place:

2C6H14 + 19O2 ==> 12CO2 + 14H2O ... balanced equation for complete combustion of hexane

Next we find the moles of hexane combusted, and then the moles of CO2.

moles hexane = 2.15 g hexane x 1 mol hexane/86 g = 0.025 moles hexane

moles of CO2 = 0.025 moles hexane x 12 moles CO2/2 moles hexane = 0.15 moles CO2

Finally, we find the volume of CO2 using the fact that at STP, 1 mole of gas = 22.4 dm³.

0.15 mol CO2 x 22.4 dm³/mol = 3.36 dm³