Correct Answer - B
(i) No. of moles of ethane = `("Mass")/("Molar mass")`
`=(0.090 kg)(30 xx 10^(-3) kg mol^(-1)) = 3` moles
The combustion of `C_(2)H_(6)` (ethane) is as-
`C_(2)H_(6)(g) + 7/2 O_(2)g to 2CO_(2)(g) + 3H_(2)O(l)`
`Deltan = sum n_(g) ("product") - sum n(g)" reactants"`
`=(2) - (3.5+1)`
`=-2.5`
`W = -Deltan RT`
`=-(-2.5 xx 8.314 xx 300) = 6234.5` J
`~= 6.234 kJ`
(ii) For 3 moles: W =`3 xx 6.234 = 18.702 kJ`
