Urea `(H_(2)NCONH_(2))` is manufactured by passing `CO_(2) (g)` through ammonia solution followed by crystallization. `CO_(2)` for the above reaction is prepared by combustion of hydrocarbon. If combustion of 236kg of a saturated hydrocarbon `(C_(n)H_(2n+2))` produces as much `CO_(2)` as required for production of `999.6 kg` urea then molecular formula of hydrocarbon is : `C_(10)H_(22)`, `C_(12)H_(26)`, `C_(13)H_(28)`, `C_(8)H_(18)`,

Options

A. `C_(10)H_(22)`
B. `C_(12)H_(26)`
C. `C_(13)H_(28)`
D. `C_(8)H_(18)`

Answer

B

Explanation

`2NH_(3)+CO_(2)rarrNH_(2)-CO-NH_(2)+H_(2)O`
moles of urea `=(999.6xx10^(3))/(60)=16.66xx10^(3)" "therefore " moles of "CO_(2)=16.66xx10^(3)`
`C_(n)H_(2n+2)+((3n+1)/(2))O_(2) to nCO_(2)+ (n+1)H_(2)O`
`(16.66xx10^(3))/(n)xx[12n+(2n+2)]=236xx10^(3)`
`n cong 12" "therefore` Molecular formula of hydrogen `=C_(12)H_(26)`.