Question

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 1

`C_(7)H_(16)+11O_(2)to7CO_(2)+8H_(2)O`
11 mol of `O_(2)` is required to burn 1 mol of `C_(7)H_(16)`.
therefore, mole of `O_(2)` required to burn 10 g of `C_(7)H_(16)`
`(1)/(10)xx11=1.1mol`
Volume of `O_(2)=PV=nRT`
`thereforeV=(nRT)/(P)=13.5L`