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


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22.4 dm3=22.4 L, the molar volume of a gas in ~ 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 mole C6H14=0.15 moles CO2. In ~ STP, 1 mole gas occupies 22.4L (22.4 dm3). Therefore 0.15 mole X 22.4 dm3/mole=3.36 dm3.


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

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

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

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

Finally, we uncover the volume that CO2 using the truth that in ~ STP, 1 mole that gas = 22.4 dm3.

0.15 mol CO2 x 22.4 dm3/mol = 3.36 dm3


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