Calculate the no. of moles of Ethane required to produce 44 g of `CO_(2(g))` after combustion .

Explanation

Balanced equation for the combustion of ethane
` C_(2) H_(6) +(7)/(2) O_(2) to 2 CO_(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
`:.` To produce 1 mole (44g) of `CO_(2)` required number of moles of ethane
` = (2 "mol ethane")/(4cancel(molCO_(2)))xx 1 cancel(molCO_(2))`
` = (1)/(2)` mole of ethane ` 0 . 5 ` mole of ethane.