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) 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.

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0.025 moles hexane X 12 moles CO2/2 moles C6H14=0.15 moles CO2. At STP, 1 mole gas occupies 22.4L (22.4 dm3). so 0.15 moles X 22.4 dm3/mole=3.36 dm3. First, we must write the correctly balanced equation for the reaction taking place:

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

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