Question

Isobutane `(C_(4)H_(10))` undergoes combustion in oxygen according to the following reaction:
`2C_(4)H_(10)(g)+13O_(2)(g) rarr 8CO_(2)(g)+10H_(2)O(e)`
When `10.00 L` of isobutane is burnt at `27^(@)C` and `1 bar` pressure, calculate the volume of `CO_(2)` produced at `120^(@)C` and `4.0 bar` pressure.

Answer 1

Step 1 Calculation of volume of `CO_(2)` at `27^(@)C` and `1"bar"` From the chemical equations, it is clear that
`2L` of `C_(4)H_(10)` produces `CO_(2)=8L`
(similar conditions of `T` and `P`)
`:. 10 L` of `C_(4)H_(10)` produces `CO_(2)=(8)/(2)xx10=40L`
(at `27^(@)C`, `1 "bar"`)
Step II Conversion of volume of `CO_(2)` at `120^(@)C` and `4.0`bar:
Here `P_(1)=1"bar"`, `P_(2)=4 "bar"`
`V_(1)=40L`, `V_(2)=?`
`T_(1)=300 K`, `T_(2)=120+273=393 K`
Using the gas equation, we get
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
`:. V_(2)=(P_(1)V_(1)T_(2))/(T_(1)P_(2))=(1xx40xx393)/(300xx4.0)=13.1 L`