2-Fluoropentane can be dehalogenated through a series of two reactions. The first reaction occurs with 20.0% excess sodium hydroxide within the solvent methanol, and produces two alkenes; 70% is 1-pentene and 30% is 2-pentene. The first reaction is shown here. CH3CFHCH2CH2CH3+CH3OH+NaOHCH2=CHCH2CH2CH3(70%)+CH3CH=CHCH2CH3(30%)+CH3OH+NaF+H2O 100% of the 2-fluoropentane reacts to form one or the other pentene product. The product stream proceeds to a separator that removes the methanol, sodium fluoride, water and unreacted sodium hydroxide. In the second reactor, 20.0% excess hydrogen gas is added to the pentene stream, and 70.0% of each pentene is converted to pentane. CH2=CHCH2CH2CH3+H2C5H12CH3CH=CHCH2CH3+H2C5H12 Given the feed concentrations in the flow diagram, if 20.0mol/hr of pentane is required, what are the flow rates of each feed and product stream? q1= mol/hr q3= mol/hrq4= mol/hr qs=