A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions What is the ratio of molecular weight to emperical formula weight of the hydrocarbon?

A gaseous hydrocarbon consumed 5 times its volume of oxygen as for com

1.	A gaseous hydrocarbon consumed 5 times its volume of oxygen as for combustion. The volume of CO_(2) produced in the reaction is thrice the volume of hydrocarbon under the same conditions What is the ratio of molecular weight to emperical formula weight of the hydrocarbon?
Answer» 1 2 3 4 Solution :`C_(2)H_(2):(24)/(26)xx100=92.3, C_(6)H_(6):(72)/(78)xx100=92.3`