1.

A gaseous mixture of ethanen, ethane and propane having total volume 200ml is subjected to combustion in excess of oxygen. Percentage of propane in original mixture is 10% then calculate volume of CO_(2)(g) obtained at same temperature and pressure.

Answer»

360 ml
390ml
420ml
can't be determined

Solution :`C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O`
`C_(2)H_(6)+(7)/(2)O_(2)rarr2CO_(2)+3H_(2)O`
`C_(3)H_(8)+5O_(2)rarr3CO_(2)+4H_(2)O`
Total volume of mixture 200ML
Volume of `C_(3)H_(8)=10%` of 200ml=20ml
so `CO_(2)` PRODUCED from propane `20xx3=60ml`
Rest of the gases =180 ml
and both produce double volume of `CO_(2)`
so `CO_(2)` produced `=180xx2=360 ml`
Total `CO_(2)(G)=360+60=420ml`


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