How many moles of ethane is equaired to produce 44 g of `CO_(2(g))` after combustion.

Explanation

Balanced equation for the combination of ethane `C_(2)H_(6)+(7)/(2)O_(2)to 2CO_(2)+3H_(2)O`
`rArr 2C_(2)H_(6)+7O_(2)to 4CO_(2)+6H_(2)O`
To produce 4 moles of `CO_(2),2` moles of ethane is required
`therefore` To produce 1 mole (44g) of `CO_2` required number of moles of ehtane
`=(2 "mol ehtane")/(4 cancel ("mol"CO_2))xx1cancel("mol"CO_2)`
`=(1)/(2)` mole of ehtane
`0.5` mole of ehtane.