What volume of `O_(2)` at 2.00 atm pressure and `27^(@)C` is requried to burn 10.0 g of heptane `(C_(7)H_(16))` ?
`C_(7)H_(16) + 11 O_(2) to 7 CO_(2) + 8 H_(2)O`

Answer

`13.71 L`

Explanation

Step I. Volume of `O_(2)` at N.T.P.
The chemical equation for the reaction is :
`{:(C_(7)H_(16),+,11O_(2),rarr7CO_(2)+8H_(2)O),(7xx12+16xx1,,11xx22.4,),(=100g,,-246.4L,):}`
100 g of `C_(7)H_(16)` require `O_(2) = 246.4 L , 10 g` of `C_(7)H_(16)` require `O_(2) = (246.4 xx 10)/(100) = 24.64 L`
Volume of `O_(2)` under exprimental conditions
`{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=24.64L,,V_(2)=?),(P_(1)=1.013"bar",,P_(2)=2 "bar"),(T_(1)=273 K,,T_(2)=27+273=300K):}`
By applying gas equation :
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1)P_(2)) = ((1.013 "bar") xx (24.64 L) xx (300 K))/((273 K) xx (2 "bar")) = 13.71 L`